College Algebra

College Algebra
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C OLLEGE A LGEBRA
Richard W. Beveridge
September 2, 2018
2018
c Richard W. Beveridge
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Contents
1 Algebra Review 5
1.1 Algebraic Simplification . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3 Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.4 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
1.5 Quadratic Equations with Complex Roots . . . . . . . . . . . . . . . 43
1.6 Multiplying and Dividing Rational Expressions . . . . . . . . . . . 50
1.7 Adding and Subtracting Rational Expressions . . . . . . . . . . . . 58
1.8 Complex Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
1.9 Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
2 Polynomial and Rational Functions 83
2.1 Representing Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . 83
2.2 Solution by Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
2.3 Solution of Polynomial Inequalities by Graphing . . . . . . . . . . . 92
3
4 CONTENTS
2.4 Solution of Rational Inequalities by Graphing . . . . . . . . . . . . . 100
2.5 Finding Factors from Roots . . . . . . . . . . . . . . . . . . . . . . . 108
2.6 Polynomial Long Division . . . . . . . . . . . . . . . . . . . . . . . . 111
2.7 Synthetic Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
2.8 Roots and Factorization of Polynomials . . . . . . . . . . . . . . . . 128
3 Exponents and Logarithms 135
3.1 Exponential and Logistic Applications . . . . . . . . . . . . . . . . . 135
3.2 Logarithmic Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
3.3 Solving Exponential Equations . . . . . . . . . . . . . . . . . . . . . 161
3.4 Solving Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . 168
3.5 Applications of the Negative Exponential Function . . . . . . . . . 175
Answer Key 185

Chapter 1
Algebra Review
This College Algebra text will cover a combination of classical algebra and ana-
lytic geometry, with an introduction to the transcendental exponential and loga-
rithmic functions. If mathematics is the language of science, then algebra is the
grammar of that language. Like grammar, algebra provides a structure to math-
ematical notation, in addition to its uses in problem solving and its ability to
change the appearance of an expression without changing the value.
1.1 Algebraic Simplification
When algebraic techniques are presented as skills in isolation, they are much sim-
pler to understand and practice. However the problem solving process in any
context involves deciding which skills to use when. Most College Algebra stu-
dents will have practiced problems in the form:
(x + 7)(x − 2) =?
or
(2x + 1)2 =?
The problems in this section deal with a combination of these processes which
are often encountered as parts of more complex problems.
5
6 CHAPTER 1. ALGEBRA REVIEW
Examples
Simplify:
3(x − 1)(2x + 5) − (x + 4)2
In this example, the simplification involves two expressions: 3(x − 1)(2x + 5) and
(x + 4)2 . The (x + 4)2 is preceded by a negative (or subtraction) sign. This text-
book will often treat −x and +(−x) as equivalent statements, since subtraction is
defined as the addition of a negative.
We will simplify each expression separately and then look to combine like terms.
3(x − 1)(2x + 5) − (x + 4)2 = 3(2x2 + 3x − 5) − (x2 + 8x + 16)
Notice that the results of both multiplications remain inside of parentheses. This
is because each one has something that must be distributed.
In the case of (2x2 + 3x − 5), there is a 3 which must be distributed, resulting in

6x2 + 9x − 15. In the case of (x2 + 8x + 16) there is a negative sign or −1 which
must be distributed, resulting in −x2 − 8x − 16. It is important in these situations
that the negative sign be distributed to all terms in the parentheses.
So:
3(x − 1)(2x + 5) − (x + 4)2 = 3(2x2 + 3x − 5) − (x2 + 8x + 16)

= 6x2 + 9x − 15 − x2 − 8x − 16
= 5x2 + x − 31
Simplify:
2(x + 3)2 − 4(3x − 1)(x + 2)
This example shows some of the same processes as the previous example. There
are again two expressions that must be simplified, each of which has a coefficient
that must be distributed. It is often helpful to wait until after multiplying the
binomials before distributing the coefficient. However, as is often true in math-
1.1. ALGEBRAIC SIMPLIFICATION 7
ematics, there are several different approaches that may be taken in simplifying
this problem.
If someone prefers to first distribute the coefficient before multiplying the bino-
mials, then the coefficient must only be distributed to ONE of the binomials, but
not both. For example, in multiplying 3∗2∗5 = 30, we can first multiply 2∗5 = 10
and then 3 ∗ 10 = 30. Each factor is multiplied only once.
In the example above we can proceed as we did with the previous example:
2(x + 3)2 − 4(3x − 1)(x + 2) = 2(x2 + 6x + 9) − 4(3x2 + 5x − 2)

= 2x2 + 12x + 18 − 12x2 − 20x + 8
= −10x2 − 8x + 26
Or, we can choose to distribute the 4 first:
2(x + 3)2 − 4(3x − 1)(x + 2) = 2(x2 + 6x + 9) − (12x − 4)(x + 2)

= 2x2 + 12x + 18 − (12x2 + 20x − 8)
= 2x2 + 12x + 18 − 12x2 − 20x + 8
= −10x2 − 8x + 26
Or, we can distribute the 4 as a negative. If we do this, then the sign in front of
the parentheses will be positive:
2(x + 3)2 − 4(3x − 1)(x + 2) = 2(x2 + 6x + 9) + (−12x + 4)(x + 2)

= 2x2 + 12x + 18 + (−12x2 − 20x + 8)
= −10x2 − 8x + 26
Distributing the 2 in front of the squared binomial must also be handled carefully
if you choose to do this. If you distribute the 2 before squaring the (x + 3), then
the 2 will be squared as well. If you choose to distribute the 2, the (x + 3)2 must
be written out as (x + 3)(x + 3):
2(x + 3)2 − 4(3x − 1)(x + 2) = 2(x + 3)(x + 3) − 4(3x − 1)(x + 2)

= (2x + 6)(x + 3) − 4(3x2 + 5x − 2)
= 2x2 + 12x + 18 − 12x2 − 20x + 8
= −10x2 − 8x + 26
Most examples in this text will distribute the coefficients as the last step before
combining like terms for a final answer.
Simplify:
3x[5 − (2x + 7)] + (3x − 2)2 − (x − 5)(x + 4)
This example has three expressions that should be simplified separately before
combining like terms. In the first expression 3x[5 − (2x + 7)], we should simplify
inside the brackets before distributing the 3x.
3x[5 − (2x + 7)] + (3x − 2)2 − (x − 5)(x + 4)
= 3x[5 − 2x − 7] + (3x − 2)2 − (x − 5)(x + 4)

= 3x[−2x − 2] + (3x − 2)(3x − 2) − (x2 − x − 20)
= −6x2 − 6x + (9x2 − 12x + 4) − x2 + x + 20
= −6x2 − 6x + 9x2 − 12x + 4 − x2 + x + 20
= 2x2 − 17x + 24
1.1. ALGEBRAIC SIMPLIFICATION 9
Exercises 1.1
Simplify each expression.
1) (x − 2)[2x − 2(3 + x)] − (x + 5)2
2) 3x2 − [7x − 2(2x − 1)(3 − x)]
3) (a + b)2 − (a + b)(a − b) − [a(2b − 2) − (b2 − 2a)]
4) 5x − 3(x − 2)(x + 7) + 3(x − 2)2
5) (m + 3)(m − 1) − (m − 2)2 + 4
6) (a − 1)(a − 2) − (a − 2)(a − 3) + (a − 3)(a − 4)
7) 2a2 − 3(a + 1)(a − 2) − [7 − (a − 1)]2
8) 2(x − 5)(3x + 1) − (2x − 1)2
9) 6y + (3y + 1)(y + 2) − (y − 3)(y − 8)
10) 6x − 4(x + 10)(x − 1) + (x + 1)2

1.2 Factoring
This section will review three of the most common types of factoring - factoring
out a Greatest Common Factor, Trinomial Factoring and factoring a Difference of
Squares.
Greatest Common Factor
Factoring out a greatest common factor essentially undoes the distributive multi-
plication that often occurs in mathematical expressions. This factor may be mono-
mial or polynomial, but in these examples, we will explore monomial common
factors.
In multiplying 3xy 2 (5x−2y) = 15x2 y 2 −6xy 3 the monomial term 3xy 2 is multiplied
or distributed to both terms inside the parentheses. The process of factorization
undoes this multiplication.
Example:
Factor 7x2 + 14x
This expression has two terms. The coefficients share a common factor of 7 and
the only variable involved in this expression is x. The highest power of the vari-
able that is shared by both terms is x1 , so this is the power of x that can be factored
out of both terms. The greatest common factor is 7x.
7x2 + 14x = 7x(x + 2)
It isn’t necessary to find the greatest common factor right away. In more compli-
cated problems, the factoring can be accomplished in pieces, similar in fashion to
reducing fractions.
1.2. FACTORING 11
Example:
Factor 42x2 y 6 + 98xy 3 − 210x3 y 2
This expression has three terms. It’s not immediately clear what the greatest
common factor of the coefficients is, but they’re all even numbers, so we could at
least divide them all by 2. The 98xy 3 term has an x1 , which means that this is the
hightest power of x that we could factor out of all the terms. The 210x3 y 2 has a
y 2 , which is the highest power of y that can be factored out of all the terms. So we
can at least proceed with these factors:
42x2 y 6 + 98xy 3 − 210x3 y 2 = 2xy 2 ∗ 21xy 4 + 2xy 2 ∗ 49y − 2xy 2 ∗ 105x2

= 2xy 2 (21xy 4 + 49y − 105x2 )
Now, we didn’t try very hard to find the greatest common factor in the beginning
of this problem, so it’s important that we continue to question whether or not
there are any remaining common factors. The 21 and 49 clearly share a common
factor of 7, so it would make sense to see if 105 is divisible by 7 as well. If we
divide 105 by 7, we see that 105 = 7*15. So, we can also factor out a common
factor of 7 from the remaining terms in the parentheses.
2xy 2 (21xy 4 + 49y − 105x2 ) = 2xy 2 (7 ∗ 3xy 4 + 7 ∗ 7y − 7 ∗ 15x2 )

= 7 ∗ 2xy 2 (3xy 4 + 7y − 15x2 )
= 14xy 2 (3xy 4 + 7y − 15x2 )
Trinomial Factoring (a = 1)
Trinomial factoring undoes the multiplication of two binomials, and it comes in
two flavors - simple and complex. The simplest form of trinomial factoring in-
volves a trinomial expression in the form ax2 + bx + c in which the value of a is 1.
This makes the task of factorization simpler than if the value of a is not 1.
Example
Factor x2 + 7x + 10
In this example, the value of a is 1, which makes this type of trinomial factoring
a little less difficult that it would otherwise be. Whether or not the value of a is 1,
the fundamental issue that governs this type of factoring is the + or − sign of the
constant term. In this problem, the constant term is positive. That means that we
need to find factors of 10 that add up to 7. This is relatively straightforward:
x2 + 7x + 10 = (x + 2)(x + 5)
A companion problem to this one is x2 − 7x + 10. Notice that, in this case, the
sign of the constant term is still positive, which means that we still need factors
of 10 that add up to 7. This means we still need to use 2 and 5. However, in
this case, instead of the +10 being produced from a multiplication of (+2)(+5),
it is the result of multiplying (−2)(−5). This is what makes the 7 in the second
example negative:
x2 − 7x + 10 = (x − 2)(x − 5)
Example
Factor x2 + 3x − 10
In this case, the sign of the constant term is negative. That means that we need to
find factors of 10 that have a difference of 3. This is still 5 and 2.
1.2. FACTORING 13
x2 + 3x − 10 = (x − 2)(x + 5)
The multiplication of the (−2) and the (+5) produce the (−10) and the fact that
the 2 and 5 have opposite signs creates the difference that gives us (+3). A com-
panion problem to this one is x2 − 3x − 10. In this case, the sign of the constant
term is still negative, which means that we still need factors of 10 that have a
difference of 3. This means we still need to use 2 and 5. However in this case,
instead of the (+3) as the coefficient of the middle term, we’ll need a (−3). To do
this we simply reverse the signs of the 2 and 5 from the previous problem:
x2 − 3x − 10 = (x + 2)(x − 5)
Now the (+2)(−5) gives us (−10), but the +2x − 5x gives us (−3x) instead of
(+3x).
Example
Factor x2 + 11x − 42
In this problem, the sign of the constant term is negative. That means that we
need factors of 42 that have a difference of 11. A systematic exploration of all the
factor pairs of 42 can help us to find the correct pair:
1 42
2 21
3 14
6 7
Here, we can see that the factors 3 and 14 have a difference of 11. This means that
we will use these factors in our answer: (x 3)(x 14). In determining how
to place the + and − signs in the parentheses, we can refer back to the original
problem: x2 + 11x − 42. If we want a difference of (+11x), then we’ll need to have
a (+14) and a (−3):
x2 + 11x − 42 = (x − 3)(x + 14)
Example
Factor x2 + 28x + 96
In this problem, the sign of the constant term is positive. That means that we
need factors of 96 that add up to 28. A systematic exploration of all the factor
pairs of 96 can help us to find the correct pair:
1 96
2 48
3 32
4 24
6 16
8 12
Here, we can see that the factors 4 and 24 add up to 28. This means that we will
use these factors in our answer: (x 4)(x 24). In determining how to place
the + and − signs in the parentheses, we can refer back to the original problem:
x2 + 28x + 96. If we want 4 and 24 to add up to (+28), then they should both be
positive:
x2 + 28x + 96 = (x + 4)(x + 24)

1.2. FACTORING 15
In building the charts of factor pairs in the previous two problems, nothing more
difficult than dividing the constant term by the numbers 1, 2, 3, 4, 5, 6, ... and so
on can help you to find the full list of factor pairs. If you don’t get a whole num-
ber when dividing - for instance 96 ÷ 5 = 19.2, then this number is not included
in the list of factor pairs.
Trinomial Factoring (a 6= 1)
If the value of a is not 1, this means that, if the trinomial is factorable, at least one
of its binomial factors also has a coefficient other than 1. For instance:
(2x + 7)(x − 3) = 2x2 + 1x − 21
If we were to try undo this multiplication through the process of trinomial fac-
toring, we should look to the sign of the constant term. In this example, the sign
is negative. This still means that we will need to find factor pairs that produce
a difference of (+1x) as the middle term. However, in this scenario, it is not just
the factors of 21 that are involved in producing the (+1x), but the combination
of the factors of 21 and the factors of the leading coefficient 2. The middle term
(+1x) comes from the multiplication of the (2x)(−3) and the multiplication of
(+7)(+1x):
(2x + 7)(x − 3)
(2x + 7)(x − 3) = 2x2 − 6x + 7x − 21

= 2x2 + x − 21
In trying to factor a trinomial like 2x2 + x − 21, we need to take this into consid-
eration. For example, if we were to factor 3x2 − 10x + 8, we should first still look
to the sign of the constant term, which, in this case, is positive. That means we
want factor pairs that will add up to 10. But we have to take into consideration
the interaction of the factors of the 3 with the factors of the 8. The 3 is a prime
number, which means that we don’t have a choice - it can only be split up into
3 ∗ 1, so we can start:
Factor 3x2 + 10x + 8

(3x ?)(x ?)
Our options for filling in the question marks will come from the factors of 8, either
8 ∗ 1 or 4 ∗ 2. The process is by trial and error:
(3x 8)(x 1) (3x 1)(x 8)
3x + 8x = 11x 24x + 1x = 25x
(3x 4)(x 2) (3x 2)(x 4)
6x + 4x = 10x 12x + 2x = 14x
We can see that the choice above:
(3x 4)(x 2)
gives us the required 10x as the middle term. Since the original problem was
3x2 + 10x + 8 we’ll want to fill in the signs as both positive:
3x2 + 10x + 8 = (3x + 4)(x + 2)
A second method for handling this type of factoring depends on how the factors
of the leading coefficient and constant term interact with each other to produce
the middle term. In this process, given the problem 3x2 + 10 + 8, we can multiply
the first and last coefficient and then look at the factor pairs of the product:
3 ∗ 8 = 24
1.2. FACTORING 17
1 24
2 12
3 8
4 6
We can see that the factor pair of 24 that adds up to 10 is 6 ∗ 4. We proceed by

splitting the 10x into 6x + 4x and then factor by grouping. If you are uncom-
fortable with factoring by grouping, then this is probably not a good method to
try. However, if you are comfortable with factoring by grouping, the rest of the
process is relatively straightforward:
3x2 + 10x + 8 = 3x2 + 6x + 4x + 8
We then factor a common factor from the first two terms and the last two terms
separately, and then factor out the common binomial factor of (x + 2):
3x2 + 10x + 8 = 3x2 + 6x + 4x + 8

= 3x(x + 2) + 4(x + 2)
= (x + 2)(3x + 4)
Example
Factor 7x2 − 5x − 18
In this example, the sign of the constant term is negative, which means that we’ll
need factor pairs that produce a difference of 5. The leading coefficient is 7, which
is prime, so, again, the only way to split up the 7 is 7 ∗ 1.
(7x ?)(x ?)
The options for filling in the question marks come from the factors of 18, for
which there are three possibilities: 18 ∗ 1, 9 ∗ 2, or 6 ∗ 3. We’ll try each of these

factor pairs in place of the question marks:
(7x 18)(x 1) (7x 1)(x 18)
18x − 7x = 11x 126x − 1x = 125x
(7x 9)(x 2) (7x 2)(x 9)
14x − 9x = 5x 63x − 2x = 61x
(7x 6)(x 3) (7x 3)(x 6)
21x − 6x = 15x 42x − 3x = 39x
The choice above:
(7x 9)(x 2)
gives us the required 5x as the middle term. Since we’re looking for a (−5x), we’ll
make the 14 negative and the 9 positive: −14x + 9x = −5x.
7x2 − 5x − 18 = (7x + 9)(x − 2)

1.2. FACTORING 19
If we want to try the other method for factoring 7x2 − 5x − 18, we would multiply
7 ∗ 18 = 126, and then work to find factor pairs of 126 that have a difference of 5:
1 126
2 63
3 42
6 21
7 18
9 14
Here, the last factor pair, 9 ∗ 14, has a difference of 5. So then we proceed to factor
by grouping:
7x2 − 5x − 18 = 7x2 + 9x − 14x − 18

= x(7x + 9) − 2(7x + 9)
= (7x + 9)(x − 2)
Notice that when the −2 was factored out from the last two terms −14x − 18, we
ended up with −2(7x + 9), because (−2) ∗ (+9) = −18. This is also important be-
cause in order to factor out the common binomial factor of (7x + 9), this binomial
must be exactly the same in both terms.
Difference of Squares
Factoring a difference of squares is actually a special form of trinomial factoring.
If we consider a trinomial of the form ax2 + bx + c, where c is a perfect square and
negative, we will find something interesting about the possible values of b that
make the trinomial factorable.
Example
Consider x2 + bx − 36
For this expression to be factorable, the middle coefficient b would need to be

equal to the difference of any of the factor pairs of 36. If we look at the possible
factor pairs, we see the following:
1 36
2 18
3 12
4 9
6 6
This means that the possible values for b that would make this expression fac-
torable are:
36 − 1 = 35 → x2 + 35x − 36 = (x + 36)(x − 1)
18 − 2 = 16 → x2 + 16x − 36 = (x + 18)(x − 2)
12 − 3 = 9 → x2 + 9x − 36 = (x + 12)(x − 3)
9 − 4 = 5 → x2 + 5x − 36 = (x + 9)(x − 4)
6 − 6 = 0 → x2 + 0x − 36 = x2 − 36 = (x + 6)(x − 6)
1.2. FACTORING 21
As we see, factoring x2 − 36 means that the factors of the perfect square 36 = 6 ∗ 6

will cancel each other out leaving 0x in the middle. If there is a perfect square as
the leading coefficient, then this number should be square rooted as well:
16x2 − 25 = (4x + 5)(4x − 5)
In the example above, the +20x and −20x as the middle terms cancel each other
out leaving just 16x2 − 25.
These three types of factoring can also be combined with each other as we see in
the following examples.
Example
Factor 2x2 − 50
This is not a trinomial because it doesn’t have three terms. It is also not a dif-
ference of squares because 2 and 50 are not perfect squares. However, there is a
common factor of 2 which we can factor out:
2x2 − 50 = 2(x2 − 25)
The expression inside the parentheses is a difference of squares and should be

factored:
2x2 − 50 = 2(x2 − 25) = 2(x + 5)(x − 5)
Example
Factor 24 − 2x − x2
Here the sign of the x2 term is negative. For this problem we can factor out
a −1 and proceed as we did with the previous problems in which the leading
coefficient was positive or we can factor it as it is:
24 − 2x − x2 = −(x2 + 2x − 24) = −(x + 6)(x − 4)

If we want to factor it as it is, we should be aware that the constant term is positive
and the quadratic term is negative, which means that we will want the factors of
24 to have a difference of 2.
24 − 2x − x2 = (6 + x)(4 − x)
Example
Factor 6x2 + 12x + 6
First, we notice that this expression has a common factor of 6. If we factor out the
6, then we should be left with an easier problem:
6x2 + 12x + 6 = 6(x2 + 2x + 1) = 6(x + 1)(x + 1) = 6(x + 1)2

1.2. FACTORING 23
Exercises 1.2
Factor each expression completely.
1) 8a2 b3 + 24a2 b2 2) 19x2 y − 38x2 y 3
3) 13t8 + 26t4 − 39t2 4) 5y 5 + 25y 4 − 20y 3
5) 45m4 n5 + 36mn6 + 81m2 n3 6) 125x3 y 5 + 60x4 y 4 − 85x5 y 2
Factor each trinomial into the product of two binomials.
7) a2 + 3a + 2 8) y 2 − 8y − 48 9) x2 − 6x − 27
10) t2 − 13t + 42 11) m2 + 3m − 54 12) x2 + 11x + 24
Factor completely. Remember to look first for a common factor. If the polynomial
is prime, state this.
13) a2 − 9 14) y 2 − 121 15) −49 + k 2
16) −64 + t2 17) 6x2 − 54 18) 25y 2 − 4
19) 200 − 2a2 20) 3m2 − 12 21) 98 − 8k 2
22) −80w2 + 45 23) 5y 2 − 80 24) −4a2 + 64

25) 8y 2 − 98 26) 24a2 − 54 27) 36k − 49k 3
28) 16y − 81y 3
Factor each trinomial completely. Remember to look first for a common factor. If
the polynomial is prime, state this.
29) 3y 2 − 15y + 16 30) 8a2 − 14a + 3 31) 9x2 − 18x + 8
32) 6a2 − 17a + 12 33) 2x2 + 7x + 6 34) 2m2 + 13m − 18
35) 20y 2 + 22y + 6 36) 36x2 + 81x + 45 37) 24a2 − 42a + 9
38) 48x2 − 74x − 10
Factor each expression completely.
39) 30 + 7y − y 2 40) 45 + 4a − a2 41) 24 − 10x − x2
42) 36 − 9x − x2 43) 84 − 8x − x2 44) 72 − 6a − a2
45) 6y 2 + 24y + 15 46) 10y 2 − 75y + 35 47) 20ax2 − 36ax − 8a

1.3. QUADRATIC EQUATIONS 25
1.3 Quadratic Equations
Quadratic equations are equations of the second degree. The solution of quadratic
equations has a long history in mathematics going back several thousand years
to the geometric solutions produced by the Babylonian culture. The Indian math-
ematician Brahmagupta used ”rhetorical algebra” (algebra written out in words)
in the 7th century to produce solutions to quadratic equations and Arab mathe-
maticians of 9th and 10th centuries followed simlar methods. Leonardo of Pisa,
also known as Fibonacci included information on the Arab approach to solving
quadratic equations in his book Liber Abaci, published in 1202.
The quadratic formula is generally used to solve quadratic equations in standard

form: ax2 + bx + c = 0. The solutions for this are:
√
−b ± b2 − 4ac
x=
2a
Now, the question is - why does this formula give solutions to the standard
quadratic equation? We can proceed as we normally do in solving linear equa-
tions - that is, by getting the x by itself. The only problem here is that instead of
just x, there are also terms involving x2 . This is where the process of completing
the square comes in handy.
We can begin with the quadratic equation in standard form:
ax2 + bx + c = 0
Just as it is easier to factor a quadratic trinomial if the leading coefficient is 1, this

process of completing the square is also easier if the leading coefficient is 1. So,
next we will divide through on both sides of this equation by a.
ax2 + bx + c 0
=
a a
ax2 bx c 0
+ + =
a a a a
x2 + ab x + c
a
=0
c
Then, we will move the to the other side of the equation to clear out some room
a
for completing the square:
b c
x2 + x + = 0
a a
c c
− =−
a a
b c
x2 + x =−
a a
Now we need to complete the square. If you are already familiar with this pro-
cess, you may wish to skip the following explanation.
If we look at what happens when we square a binomial like (x+3)2 , we will begin
to notice a pattern.
(x + 3)2 = (x + 3)(x + 3) = x2 + 6x + 9
(x + 4)2 = (x + 4)(x + 4) = x2 + 8x + 16
(x + 5)2 = (x + 5)(x + 5) = x2 + 10x + 25
(x + 6)2 = (x + 6)(x + 6) = x2 + 12x + 36
Our goal in the derivation of the Quadratic Formula is to rewrite the expression
x2 + ab x as a perfect square in the form (x+ )2 . The reason that we want to do
this is that writing an expression as a binomial squared eliminates the problem of
having both an x and an x2 , which was preventing us from getting the x by itself
in the standard quadratic equation.
If we can figure out what should take the place of the blanks in the statement:
x2 + ab x+ = (x+ )2
then we will be well on our way to deriving the quadratic formula.

If we re-examine the sample perfect binomial squares from the previous page, we
note a useful pattern. This is that the blank in the parentheses (x+ )2 is filled
by a number that is one-half the value of the linear coefficient - or the coefficient
of the x1 term. Notice that in x2 + 6x + 9 = (x + 3)2 , 3 is half of 6, in x2 + 8x + 16 =
(x + 4)2 , the 4 is half of 8, and so on. If we want to write x2 + ab x+ as a perfect
2
square in the form (x+ ) , the blank in the parentheses should be filled by:
1 b b
∗ =
2 a 2a
2
b b
Now, it’s not true that x2 + x+ = x+
a 2a
We’re missing the constant term on the left. However, if we return to our perfect
square examples, we can see that the constant term is always the square of the
term inside inside the parentheses. So, we can restate our problem now as:
2 2
2b b b
x + x+ = x+
a 2a 2a
2
b2

2 b b
x + x+ 2 = x+
a 4a 2a
So, if we return to our originial problem, we were saying that:
b c
x2 + x + = 0
a a
c c
− =−
a a
b c
x2 + x =−
a a
b2
We can add 2 to both sides of this equation and then restate the left hand side
4a
as a perfect square of a binomial:
b c
x2 + x =−
a a
b2 b2
+ 2 =+ 2
4a 4a
2b b2 c b2
x + x+ 2 =− + 2
a 4a a 4a
2
b2

b c
x+ =− + 2
2a a 4a
The last tricky bit of this derivation is adding the two fractions on the right hand
side. The common denominator for these fractions is 4a2 , so we’ll need to multi-
c 4a 4ac b2 − 4ac
ply the − by to get − 2 . Then the right hand side will be :
a 4a 4a 4a2
2
b2 − 4ac

b
x+ =
2a 4a2
Then, we can take the square root of both sides and get the x by itself:
s 2 r
b b2 − 4ac
x+ =±
2a 4a2
√
b ± b2 − 4ac
x+ = √
2a 4a2
√
b ± b2 − 4ac
x+ =
2a 2a
b
Subtracting from both sides is easy since we already have a common denom-
2a
inator:
√
b ± b2 − 4ac
x+ =
2a 2a
b b
− =−
2a 2a
√
−b ± b2 − 4ac
x=
2a
At the college algebra level, it is often useful to program the quadratic formula
onto a graphing calculator both for easy use and also to learn a little bit about
programming. The following program is a simple example of this for the TI-
84 series of graphing calculators. Graphing calculators also often have built-in
polynomial solver feature that can be used to solve quadratics.
Press the ”prgm” key in the top middle of the calculator keypad. This will bring
up a screen that shows EXEC EDIT NEW across the top. Arrow over across
the top to ”NEW,” and then select 1: Create New.
This will bring up a screen asking you to name the program. You should see
PROGRAM and then underneath it, ”Name=”. The alpha lock is on automat-
ically, so any key you press will type the letter associated with it. Name your
program and press ENTER. You should see PROGRAM: Name, with whatever
name you’ve chosen for your program. Underneath this you will se a colon :.
This is where you will enter the commands for the program.
First we need to enter the values for A, B and C from the quadratic equation into
the calculator. To do this, press the ”prgm” key again. Across the top of the screen
you should see CTL I/O COLOR EXEC.
Arrow over to I/O. This is the ”input/output” menu. Choose number 2:Prompt.
This will return you to the program screen where you will see :Prompt under the
name of the program. After :Prompt, type A, B, C. You’ll need to use the ”alpha”
key to access the letters and the comma is right above the 7 key.
PROGRAM: Name (whatever name you’ve chosen should show here)

:Prompt A, B, C
On the next line of the program, we will take the values of A, B and C and use
them to calculate the values of the roots of the equation. Type in the following:
PROGRAM: Name
:Promptp
A, B, C
:(−B + p(B 2 − 4AC))/(2A) → R
:(−B − (B 2 − 4AC))/(2A) → S
In typing these two lines it’s important that when you type −B, you use the nega-
tive key next to the decimal point, rather than the subtraction key. The calculator
is very picky about this. When you’re typing the B 2 − 4AC, you’ll need to use the
subtraction key on the far right of the keypad.
Also notice the double parentheses - one set for the numerator of the fraction
and one set for the square root. If you don’t type this in correctly it will produce
wrong answers. The arrow in the formula stores the values of the answer in the
variables R and S, and the arrow is produced by the ”sto→” key just above the
ON button in the lower left of the keyboard.
Now that we’ve given the calculator the values for A, B and C and then had
the calculator find the roots of the eqatiuon, we need to display the answers. If
you press the prgm key and arrow over to the I/O menu again, you can choose
3:Disp. This will Display the answers that we’ve stored as R and S.
PROGRAM: Name
:Prompt√A, B, C
:(−B + √B 2 − 4AC))/(2A) → R
:(−B − B 2 − 4AC))/(2A) → S
:Disp R,S
Now we can test the program with some simple equations. To run the program,
press the program key and choose the program you’ve created either by selecting
it and pressing enter, or by pressing the number for the program in the list. This
should bring you back to the calculation screen, where you can run the program
by pressing enter. The calculator should then ask you for the values of A, B and
C.
Solve for x: 2x2 − x − 1 = 0
In this example the values for A, B and C are:

A=2
B = −1
C = −1
Again, it’s important that you use the negative sign key next to the decimal point
for the values of any negative coefficients and not the subtraction key. The calcu-
lator should return values of 1 and −0.5 as the solutions.
Solve for x: x2 + x + 1 = 0
In this example the values for A, B and C are:

A=1
B=1
C=1
The calculator should return values of −0.5 ± 0.8660254038i as solutions.
If you get an error message saying ”NONREAL ANSWERS,” you’ll need to ad-
just the calculator setting to allow for complex valued answers. You can do this
by presing the ”mode” key in the top left of the keypad and arrowing down to the
line that reads ”REAL a+bi reˆ(θi).” You can then arrow over to ”a+bi” and
press enter. This will allow the calculator to compute complex valued answers.
Something very important to remember about the quadratic formula is that the
equation must be in standard form in order to identify the values of A, B and C
to use in the formula. For example in the equation:
3x2 − 7 = 2x
it is important to understand that the values of A, B and C come from the stan-
dard form of the equation and not the present form of the equation. There are
several pitfalls to watch out for in this equation. First of all, the 2x is on the op-
posite side of the equation from the other terms. That means that the value of B
IS NOT +2. Also, if we were to move the 2x to the other side to put the equation
in standard form, it is not the order of the terms, but degree of the variable that
determines whether a coefficient is identified as A, B or C.
In moving the 2x to the other side of the equation, I have seen students put the
term they’re adding to that side as the last term. There is nothing wrong about
this, but if you do that, you must be careful about identifying the values of A, B
and C.
3x2 − 7 = 2x
−2x = −2x
3x2 − 7 − 2x = 0
There is nothing wrong about the way the equation above is written depsite the
fact that it is not in ”standard form.” The important thing to remember is that ”A”
is not the coefficient of whichever term is listed first. It is the coefficient of the
quradratic, or x2 term. Likewise, ”B” is not the coefficient of the second term, but
rather the coefficient of the linear, or x1 term. And ”C” is not whichever number
comes last, but rather the value of the constant term. So in the equation above,
however it is written, the value of A is +3, B is −2 and C is −7.
Exercises 1.3
Solve for x in each equation. Round any irrational values to the nearest 1000th.
1) x2 + 7x = 2 2) 5x2 − 3x = 4
3 2
3) 4
x = 78 x + 1
2
4) 2 2
3
x − 1
3
= 5
9
√ √
5) 2x2 + ( 5)x − 3 = 0 6) 3x2 + x − 2=0
7) 2.58x2 − 3.75x − 2.83 = 0 8) 3.73x2 + 9.74x + 2.34 = 0
9) 5.3x2 + 7.08x + 1.02 − 0 10) 3.04x2 + 1.35x + 1.234 = 0
11) 7x(x + 2) + 5 = 3x(x + 1) 12) 5x(x − 1) − 7 = 4x(x − 2)
13) 14(x − 4) − (x + 2) = (x + 2)(x − 4)
14) 11(x − 2) + (x − 5) = (x + 2)(x − 6)

1.4 Complex Numbers
Our number system can be subdivided in many different ways. The most basic
form of mathematics is counting and almost all human cultures have words to
represent numbers (the Pirahã of South America are a notable exception). Thus
the most basic set of numbers is the set of counting numbers represented by the
double barred N: N = {1, 2, 3, 4, 5, 6, 7, . . . } (we will set aside the debate as to
whether or not zero should be included in this set).
If we try to subtract a larger counting number from a smaller counting number

we find that there are no members in the set of counting numbers to represent
the answer in this situation. This extends the set of natural numbers to the set of
integers: Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . . }. The integers are represneted by the
double barred Z, for the German word for numbers - ”zahlen.” In the earliest ap-
pearances of negative numbers in the Chinese and Indian mathematical systems,
negative values were often used to represent debt. Because Greek mathematics
was based on Geometry, they did not use negative numbers.
Moving to multiplication and division, if we question the value of 8 ÷ 2 = 4 ver-

sus 8 ÷ 3 =?, we once again must expand our conception of numbers to allow
for an answer to the second question 8 ÷ 3 =?. Understanding ratios of whole
numbers or Rational numbers allows solutions to such problems. The set of Ra-
tional numbers isrepresented by the double barred Q, to represent a quotient:
a
Q= : a, b ∈ Z .
b
The Greek understanding of numbers mostly stopped here. They felt that all
quantities could be represented as the ratio of whole numbers. The length of
the diagonal of a square whose sides are of length 1 produced considerable con-
sternation among the Pythagoreans as a result of this. Using the Pythagorean
Theorem for the diagonal of a square whose sides √ are of length 1 shows that the
2 2 2
diagonal would be c = 1 + 1 = 2, thus c = 2. This number cannot be rep-
resented as a ratio of whole numbers. This new class of numbers adds the set
of irrational numbers to the existing set of rational numbers to create the Real
numbers, represented with a double barred R: R.
1.4. COMPLEX NUMBERS 35
This hierarchy of numbers is often represented in the following diagram:
N
One of the best ways to conceptualize the Real number system is on the number
line - every point on the number line corresponds to a unique Real number and
every Real number corresponds to a unique position on the Real number line.
-4 -3 -2 -1 0 1 2 3 4
√
2
After the development of the printing press in the 15th century, Fibonacci’s Liber
Abaci was translated into Italian from Latin and read throughout Italy. As a result,
Italy became a thriving center of mathematics until the 17th century, when the
center of European mathematics moved north to France, Germany and England.
Throughout the 1500’s Italian mathematicians such as Girolamo Cardano, Raphael

Bombelli and Niccolo Fontana Tartaglia worked to extend the ideas in Fibonacci’s
book. They produced formulas to solve cubic (x3 ), and quartic (x4 ) degree equa-
tions. In solving some of these equations they found that their formulas some-
times produced negative values under a square root. None of the known number
systems could accomodate this possibility. In Cardano’s book on algebra Ars
Magna, he encounters a problem which involves the square root of a negative
number. He says, ”It is clear that this case is impossible. Nevertheless we will
work thus...” and he proceeds to compute a valid complex solution
√ to the prob-
lem. Mathematicians eventually defined the complex unit −1 = i and then
devised a system in which all complex numbers are a combination of a Real part
(a) and an ”imaginary” part (bi).
The complex numbers are a two dimensional number system represented by the
double√barred C: C = {a + bi : a, b ∈ R}, where i is the complex unit defined
as i = −1. Throughout the late 1700’s and early 1800’s mathematicians gradu-
ally moved towards a geometrical interpretation of the two-dimensional complex
numbers. What is today known as the ”Argand diagram” represents the real val-
ued portion of a complex number along a horizontal axis and the multiple of the
complex unit along the vertical axis.
Im
5i
−1 + 4i 4i
3i
2i 3 + 2i
1i
R
−5 −4 −3 −2 −1 1 2 3 4 5
−1i
−2i
−3i
−4i
−5i
Expressing square roots of negative numbers

Square roots of negative quantities are generally expressed as a multiple of i.
Examples
√
−4 = 2i.
√
−25 = 5i
√
−7 ≈ 2.646i
Adding, subtracting and multiplying with complex numbers

Calculating with complex numbers has many similarities to working with vari-
ables. The real part and imginary part are treated separately for addition and
subtraction, but can be multiplied and divided.
Examples
Compute the following:
(6 − 4i) + (−2 + 7i) = 4 + 3i
(−9 + 2i) − (−4 + 6i) = −9 + 2i + 4 − 6i = −5 − 4i
3(10 + i) = 30 + 3i
−7i(−5 + 8i) = 35i − 56i2
Now we encounter an interesting fact about

√ complex numbers and, in particular,
the complex unit i. By definition, i = −1. Therefore, if we square i we should
get −1. In the last example problem above, we can replace the i2 with −1 to finish
the problem.
−7i(−5 + 8i) = 35i − 56i2
= 35i − 56(−1)
= 35i + 56
= 56 + 35i
Examples
Compute the following:
(8 − 5i)(1 − 4i) = 8 − 32i − 5i + 20i2
= 8 − 37i + 20(−1)
= 8 − 37i − 20
= −12 − 37i
(9 + 2i)2 = (9 + 2i)(9 + 2i)
= 81 + 18i + 18i + 4i2
= 81 + 36i + 4(−1)
= 81 + 36i − 4
= 77 + 36i
Powers of i
The powers of i follow an interesting pattern based on the definition that i2 = −1.
We can see that i1 = i and that i2 = −1, as a result, i3 = i2 ∗ i1 = −1 ∗ i = −i.
In a similar fashion, i4 = i2 ∗ i2 = (−1)(−1) = 1.
This means that i5 = i4 ∗ i = 1 ∗ i = i.
If we put all of this information together we get the following:
i1 = i
i2 = −1
i3 = −i
i4 = 1
i5 = i1 = i
i6 = i2 = −1
i7 = i3 = −i
i8 = i4 = 1
In other words, every power of i is equivalent to either i, −1, −i, or 1. To de-

termine which of these values a power of i is equivalent to, we need to find the
remainder of the exponent when it is divided by 4.
Example
Simplify i38
Since every i4 = 1, then i38 = i36 ∗ i2 = (i4 )9 ∗ i2 = 19 ∗ i2 = i2 = −1
Since 38 is 2 more than a multiple of 4, then i38 = i2 = −1.

Exercises 1.4
Im
10i
9i
8i
7i
6i
5i
4i
3i
2i
1i
R
−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10
−1i
−2i
−3i
−4i
−5i
−6i
−7i
−8i
−9i
−10i
Graph the following complex numbers.
1) 2 + 5i 2) 4 − 3i 3) −2 + 6i 4) −3 − 5i
5) 4 6) −2i 7) 7−i 8) −1 + i
9) −8 + 4i 10) 8 + 3i 11) 7i 12) −5 − 9i

Express each quantity in terms of i. Round irrational values to the nearest 1000th.
√ √ √ √
13) −36 14) −81 15) −100 16) −49
√ √ √ √
17) −4 18) −25 19) −2 20) −6
√ √ √ √
21) −10 22) −31 23) −5 24) −3
Perform the indicated operation and simplify.
25) (6 + 7i) + (5 + 3i) 26) (4 − 5i) + (3 + 9i)
27) (9 + 8i) − (1 − 2i) 28) (2 + i) − (6 − 4i)
29) (7 − 4i) − (5 − 3i) 30) (8 + i) − (4 + 3i)
31) (7i)(6i) 32) (4i)(−8i)
33) (−2i)(5i) 34) (12i)(3i)
35) (1 + i)(3 + 2i) 36) (1 + 5i)(4 + 3i)
37) (6 − 5i)(2 − 3i) 38) (8 − 3i)(2 + i)
39) (−3 + 4i)(−1 − 2i) 40) (−7 − i)(3 − 5i)
41) (4 − 2i)2 42) (−5 + i)2

43) (3 + i)(3 − i) 44) (2 + 6i)(2 − 6i)
45) (9 − 4i)(9 + 4i) 46) (5 + 2i)(5 − 2i)
Express as either i, −1, −i, or 1.
47) i3 48) i7 49) i21 50) i13
51) i29 52) i56 53) i72 54) i35
55) i66 56) i103 57) i16 58) i53
59) i11 60) i42 61) i70 62) i9

1.5. QUADRATIC EQUATIONS WITH COMPLEX ROOTS 43
1.5 Quadratic Equations with Complex Roots
In Section 1.3, we considered the solution of quadratic equations that had two
real-valued roots. This was due to the fact that in calculating the roots for each
equation, the portion of the quadratic formula that is square rooted (b2 −4ac, often
called the discriminant) was always a positive number.
For example, in using the quadratic formula to calculate the the roots of the equa-
tion x2 − 6x + 3 = 0, the discriminant is positive and we will end up with two
real-valued roots:
x2 − 6x + 3 = 0
a = 1, b = −6, c = 3
p
−(−6) ± (−6)2 − 4(1)(3)
x=
2∗1
√
6± 36 − 12
=
2
√
6± 24
=
2
6 ± 4.899
≈
2
6 + 4.899 6 − 4.899
≈ ≈
2 2
10.899 1.101
≈ ≈
2 2
≈ 5.449 ≈ 0.551
When we added and subtracted the square root of 24 to 6 in the quadratic for-
mula, this created two answers, and they were real-valued because the square
root of 24 is real-valued.
Another way to see this is graphically. If we graph y = x2 − 6x + 3 and find the x

values that make y = 0 , these will appear along the x-axis, and will be the same
values that solve the equation x2 − 6x + 3 = 0.
10
x ≈ 0.551 x ≈ 5.499
2 4 6
−5
If we consider a related, but slightly different equation to start with, these rela-
tionships between the roots, the discriminant and the graphical intersections will
be slightly different.
x2 − 6x + 9 = 0
a = 1, b = −6, c = 9
p
−(−6) ± (−6)2 − 4(1)(9)
x=
2∗1
√
6± 36 − 36
=
2
√
6± 0
=
2
6
= =3
2
Because the discriminant was 0 in this problem, we only get one real-valued an-
swer.
Graphically, the additional 6 that was added to the original equation to change it
from x2 − 6x + 3 to x2 − 6x + 9 shifts every y value on the graph up 6 units.
15
10
5 y = x2 − 6x + 9
x=3 x ≈ 5.499
x ≈ 0.551 2 4 6
−5 y = x2 − 6x + 3
If we add an additional three units to the constant term of this quadratic equation,
we encounter a third possibility.
x2 − 6x + 12 = 0
a = 1, b = −6, c = 12
p
−(−6) ± (−6)2 − 4(1)(12)
x=
2∗1
√
6± 36 − 48
=
2
√
6± −12
=
2
√
6 ± i 12
=
2
6 ± 3.464i
≈
2
6 3.464i
= ±
2 2
≈ 3 ± 1.732i
Here the discriminant is negative, which leads to two complex-valued answers.

If the equation has real-valued coefficients, the complex roots will always come
in conjugate pairs. Complex conjugates share the same real-valued part and have
opposite signs in their complex-valued (or imaginary) parts: a ± bi
Graphically, the previous problem was one step away from not intersecting the
x-axis at all and the additional three units that we added on to get y = x2 −
6x + 12 moves the graph entirely away from the x-axis. Because the roots are
complex-valued, we don’t see any roots on the x-axis. The x-axis contains only
real numbers.
15
y = x2 − 6x + 12
10
2 4 6
−5
Since the calculator has been programmed for the quadratic formula, the focus of
the problems in this section will be on putting them into standard form.
Example
Solve for x.
(2x + 1)(x + 5) − 2x(x + 7) = 5(x + 3)2
2x2 + 11x + 5 − 2x2 − 14x = 5(x + 3)(x + 3)
−3x + 5 = 5(x2 + 6x + 9)
−3x + 5 = 5x2 + 30x + 45
0 = 5x2 + 33x + 40
a = 5, b = 33, c = 40
x = −5, −1.6
The fact that the roots of this equation were rational numbers means that the
equation could have been solved by factoring.
0 = 5x2 + 33x + 40
0 = (5x + 8)(x + 5)
5x + 8 = 0 x+5=0
5x = −8 x = −5
x = −1.6
Example
Solve for x.
(x − 2)2 + 3(4x − 1)(x + 1) = 7(x + 1)(x − 1)
x2 − 4x + 4 + 3(4x2 + 3x − 1) = 7(x2 − 1)
x2 − 4x + 4 + 12x2 + 9x − 3 = 7x2 − 7
13x2 + 5x + 1 = 7x2 − 7
6x2 + 5x + 8 = 0
a = 6, b = 5, c = 8
5
x ≈ −0.416 ± 1.077i ≈ − 12 ± 1.077i
Exercises 1.5
Solve for x in each equation. Round any irrational values to the nearest 1000th.
1) 3x2 − 3x = 4 2) 4x2 − 2x = 7
3) 5x2 = 3 − 7x 4) 3x2 = 21 = 14x
5) 6x2 + 1 = 2x 6) 5x − 3x2 = 17
7) (5x − 1)(2x + 3) = 3x − 20 8) (x + 4)(3x − 1) = 9x − 5
9) (x − 2)2 = 8x(x − 1) + 10 10) (2x − 3)2 = 2x − 7x2
11) (x + 5)(x − 6) = (2x − 1)(x − 4) 12) (3x − 4)(x + 2) = (2x − 5)(x + 5)

1.6 Multiplying and Dividing Rational Expressions
Reducing Rational Expressions

A rational expression is simply an algebraic fraction, and our first consideration
will be to reduce these expressions to lowest terms in the same way that we re-
6 2
duce numerical fractions to lowest terms. When we reduce to by canceling
15 5
the common factor of three, we are removing a redundant factor of 1 in the form
3
of .
3
6 3∗2 3 2
= = ∗
15 3∗5 3 5
2
=1∗
5
2
=
5
Similarly, if there is a common factor that can be factored out of an algebraic

fraction, this also can be canceled.
21x + 14 7(3x + 2)
=
7x + 7 7(x + 1)
7 3x + 2
= ∗
7 x+1
3x + 2
=1∗
x+1
3x + 2
=
x+1
It’s important to remeber that only common factors can be canceled. This means
1.6. MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS 51
that the first priority in each problem will be to identify the factors of the numer-
ator and denominator to see if they share any common factors.
Example
Reduce to lowest terms.
x2 + 4x − 12
x2 − 4
x2 + 4x − 12 (x + 6)(x − 2)
=
x2 − 4 (x + 2)(x − 2)
(x−2)
(x + 6)
=
(x−2)
(x + 2)
x+6
=
x+2
Notice that we can’t cancel the 6 and the 2 in the final answer because they aren’t
factors. The plus signs in the numerator and denominator prevent us from can-
celling the 6 and the 2.
In the previous examples, we saw that cancelling out common factors in the nu-
merator and denominator was actually a process of eliminating a redundant fac-
tor of 1. In the following example, we’ll see a slightly different form of cancelling.
Example
Reduce to lowest terms.
16 − x2
x2 + x − 20
16 − x2 (4 + x)(4 − x)
2
=
x + x − 20 (x + 5)(x − 4)
In this problem, there are no common factors, but we can do some cancelling. We
can see that (4 − x) and (x − 4) are not the same expression. In the first binomial,
the 4 is positive and the x is negative, whereas in the second binomial, the 4 is
4−x
negative and the x is positive. So, we know that 6= 1. However, if we factor
x−4
a (−1) out of the numerator, we will see an interesting phenomenon:
4−x −1(−4 + x)
=
x−4 x−4
−1(x − 4)
=
x−4
x−4
= −1 ∗
x−4
= −1 ∗ 1 = −1
4−x 4−x
Therefore, although 6= 1, we can say that = −1. This will allow us to
x−4 x−4
cancel (4 − x) and (x − 4) and replace them with (−1).
16 − x2 (4 + x)(4 − x)
=
x2 + x − 20 (x + 5)(x − 4)
(4 + x)(4−x)

=
(x−4)(−1)
(x + 5)
In the final answer, the (−1) can be placed in the denominator or the numerator,
but not both. It can also be placed in front of the fraction.
4+x −1(4 + x)
=
−1(x + 5) x+5
4+x
=−
x+5
Multiplying and Dividing Rational Expressions

In multiplying and dividing rational expressions, it is often easier to identify and
cancel out common factors before multiplying rather than afterwards. Multiply-
ing rational expressions works the same way that multiplying numerical frac-
tions does - multiply straight across the top and straight across the bottom. As a
result, any factor in either numerator of the problem will end up in the numera-
tor of the answer. Likewise, any factor in either denominator of the problem will
end up in the denominator of the answer. Thus, any factor in either numerator
can be cancelled with any factor in either denominator.
Example
Multiply. Express your answer in simplest form.
x2 + 5x + 6 x2 − 2x − 15
∗ 2
25 − x2 x + 6x + 9
x2 + 5x + 6 x2 − 2x − 15 (x + 2)(x + 3) (x − 5)(x + 3)
2
∗ 2 = ∗
25 − x x + 6x + 9 (5 + x)(5 − x) (x + 3)(x + 3)
(x + 2)
(x+3)

(x−5)(x + 3)

= ∗

(5− x)(−1)
(5 + x)
(x+3)
(x+ 3)
x+2
=−
x+5
Dividing rational expressions works in much the same way that dividing nu-
merical fractions does. We multiply by the reciprocal. There are several ways to
demonstrate that this is a valid definition for dividing. First, it is important to
understand that the fraction bar is the same as a “divided by” symbol:
8
= 8 ÷ 2 = 4.
2
The same is true for dividing fractions:
1

1 2 3
÷ = 2
.
3 5 5
1

3
We can take the complex fraction 2
and multiply it by 1 without changing its
5
value:
1

1 2 3
÷ = 2
∗1
3 5 5
1

3
= 2
5
We can multiply by any form of 1 we want to and not change the value of the
result.
1 1

3 3
2
∗1= 2
5 5
1 1

3 9 3
2
∗ = 2
5
9 5
1 1

3 12 3
2
∗ = 2
5
12 5
With a carefully chosen form of 1, we can transform the division problem into a
multiplication problem.
1 5 1 5

3 2 3
∗ 2
2
∗ 5
= 2 5
5 2 5
∗ 2
1 5
3
∗ 2
=
1
1 5
= ∗
3 2
5
=
6
In this way, we can redefine division as multiplication by a reciprocal.
Example
Divide the expressions. Express your answer in lowest form.
2x2 − x − 3 x2 + 5x + 4
÷
x2 − x − 12 16 − x2
2x2 − x − 3 x2 + 5x + 4 2x2 − x − 3 16 − x2
÷ = ∗
x2 − x − 12 16 − x2 x2 − x − 12 x2 + 5x + 4
(2x − 3)(x + 1) (4 + x)(4 − x)

= ∗
(x − 4)(x + 3) (x + 1)(x + 4)
(2x − 3)(x+1)

(4+x)(4 − x)(−1)

= ∗

(x− 4)(x + 3)

(x+ 1)
(x + 4)

2x − 3
=−
x+3
Exercises 1.6
Reduce each expression to lowest terms.
3x + 9 4x2 + 8x
1) 2)
x2 − 9 12x + 24
x2 − 2x 15x2 + 24x
3) 4)
6 − 3x 3x2
24x2 x2 + 4x + 4
5) 6)
12x2 − 6x x2 − 4
25 − y 2 3y 2 − y − 2
7) 8)
2y 2 − 8y − 10 3y 2 + 5y + 2
x2 + 4x − 5 x − x2
9) 10)
x2 − 2x + 1 x2 + x − 2
x2 + 5x − 14 2x2 + 5x − 3
11) 12)
2−x 1 − 2x
Multiply or divide the expressions in each problem.
Express your answers in lowest terms.
3x + 6 x 4x2 7x − 28
13) 2
∗ 2 14) 2
∗
5x x −4 x − 16 6x
2a2 − 7a + 6 4a2 + 12a + 9 4a2 − 4a − 3 16a2

15) ∗ 16) ∗
4a2 − 9 a2 − a − 2 8a + 4a2 4a2 − 6a
x2 − y 2 (x + y)2 2x2 + x − 3 2x − 2
17) ∗ 18) ∗ 2
(x + y)3 (x − y)2 2
x −1 2x + 5x + 3
6x 3x − 9 12x 4x2
19) 2
÷ 20) ÷ 2
x − 4 2x + 4 5x + 20 x − 16
9x2 + 3x − 2 3x + 2 2a2 − 5a − 3 2a + 1
21) ÷ 22) ÷
6x2 − 2x 6x2 4a2 + 2a 4a
x2 + 7x + 6 x2 + 5x − 6 x2 + 7x + 10 3x2 + 7x + 2
23) ÷ 2 24) ÷
x2 + x − 6 x + 5x + 6 x2 − x − 30 9x2 − 1
2x2 − x − 28 4x2 + 16x + 7 9x2 + 3x − 2 9x2 − 6x + 1

25) ÷ 2 26) ÷
3x2 − x − 2 3x + 11x + 6 12x2 + 5x − 2 8x2 − 10x − 3
1.7 Adding and Subtracting Rational Expressions
Just as we do with numerical fractions, we will need to have common denomina-

1 1
tors in order to add or subtract algebraic fractions. When we add + , we make
2 3
a common denominator of 6 so that we can add them together.
1 1 1 3 1 2
+ = ∗ + ∗
2 3 2 3 3 2
3 2
= +
6 6
5
=
6
Because the denominators, 2 and 3, are prime and don’t share any common fac-
tors, the common denominator is simply 3 ∗ 2 = 6. We can see a similar result in
adding algebraic fractions.
Example
Add the fractions. Express your answer in lowest terms.
2 x
+
x x−3
2 x 2 x−3 x x
+ = ∗ + ∗
x x−3 x x−3 x−3 x
2(x − 3) x∗x
= +
x(x − 3) x(x − 3)
2x − 6 + x2 x2 + 2x − 6
= =
x(x − 3) x(x − 3)
1.7. ADDING AND SUBTRACTING RATIONAL EXPRESSIONS 59
It’s important to be aware that in subtraction, the negative sign representing sub-
traction must be distributed to all terms in the second numerator.
Example
Subtract the given expressions. Express your answer in lowest terms.
6 x+5
−
x+1 x−2
6 x+5 6 x−2 x+5 x+1

− = ∗ − ∗
x+1 x−2 x+1 x−2 x−2 x+1
6(x − 2) − (x + 5)(x + 1)
=
(x + 1)(x − 2)
6x − 12 − (x2 + 6x + 5)
=
(x + 1)(x − 2)
6x − 12 − x2 − 6x − 5
=
(x + 1)(x − 2)
−x2 − 17
=
(x + 1)(x − 2)
In other situations, the denominators may share a common factor. In this case,
we can turn one of the denominators into the other one:
Example
Add the given fractions.
7 2
+
x2 + 8x + 15 x + 3
7 2 7 2
+ = +
x2 + 8x + 15 x + 3 (x + 3)(x + 5) x + 3
We can turn (x + 3) into x2 + 8x + 15 by multiplying by (x + 5)
7 2 7 2 x+5
+ = + ∗
(x + 3)(x + 5) x + 3 (x + 3)(x + 5) x + 3 x + 5
7 2(x + 5)
= +
(x + 3)(x + 5) (x + 3)(x + 5)
7 + 2x + 10
=
(x + 3)(x + 5)
2x + 17
=
(x + 3)(x + 5)
Sometimes, the answer we end up with is not in lowest terms:
Example
Add the fractions.

x 8
+ 2
x + 2 x + 8x + 12
x 8 x 8
+ 2 = +
x + 2 x + 8x + 12 x + 2 (x + 2)(x + 6)
x x+6 8
= ∗ +
x + 2 x + 6 (x + 2)(x + 6)
x(x + 6) 8 x(x + 6) + 8
= + =
(x + 2)(x + 6) (x + 2)(x + 6) (x + 2)(x + 6)
x(x + 6) + 8 x2 + 6x + 8
=
(x + 2)(x + 6) (x + 2)(x + 6)
The numerator is factorable:
x2 + 6x + 8 (x + 2)(x + 4)
=
(x + 2)(x + 6) (x + 2)(x + 6)
(x+2)(x

+ 4)
=

(x+ 2)(x + 6)

x+4
=
x+6
Exercises 1.7
Add or subtract the given expressions.
1 1 3 1
1) − 2) −
x−1 x y−6 y
2 4 3 4
3) + 4) −
x−3 x+3 x+4 x−2
3 k−4 a+1 a
5) − 6) −
k+2 k+5 a a+1
2y y x 4
7) − 8) +
y 2 − 25 y − 5 x2 − 1 x + 1
1 x 9y y+1
9) + 10) −
x−3 x+1 y−4 y+5
Add or subtract the given expressions. Express your answers in lowest terms.
b b+1 4x + 1 x−3
11) − 12) +
b + 1 2b + 2 8x − 12 2x − 3
2 1 1 4
13) + 14) − 2
a2 + 4a + 3 a + 3 y + 6 y + 8y + 12
x+1 x2 x+1 x2 + 1
15) − 16) −
2x + 4 2x2 − 8 x + 2 x2 − x − 6
2x 2x x 3x + 3 1
17) + − 18) +
x2 − 3x + 2 x − 1 x − 2 2x2− x − 1 2x + 1
4a 3a 4a 2 8 − 4y
19) − + 2 20) − 2
a − 2 a − 3 a − 5a + 6 y − 3 y − 8y + 15
2x 3x x+3 a 2 3(a − 2)
21) + − 2 22) − + 2
x−1 x+1 x −1 a−1 a+2 a +a−2
x x+7 x−2 2y + 5 y−9

23) + 2 − 24) 2
− 2
x−1 x −1 x+1 y − 16 y − y − 12
1.8 Complex Fractions
Complex fractions involve simplifying a rational expression which has a compli-

cated numerator and/or denominator.
Example
Simplify.
x
3+
x+2
x+3
1−
x−1
There are a variety of ways to approach this problem. One of the most straightfor-
ward ways to simplify the expression above is to create common denominators
for the numerator and the denominator so that each one is a single fractional
expression:
x 3 x+2 x
3+ ∗ +
x+2 1 x+2 x+2
x+3
= 1 x−1 x+3
1− ∗ −
x−1 1 x−1 x−1
3x + 6 + x
x+2
=
x − 1 − (x + 3)

x−1
4x + 6
x+2
= −4 (Now this is a division problem)
x−1
4x + 6 x − 1 2(2x + 3) x − 1
= ∗ = ∗
x+2 −4 x+2 −4
2 (2x + 3) x−1 (2x + 3)(x − 1)

= ∗ =
x+2 −4(−2)
−2(x + 2)
1.8. COMPLEX FRACTIONS 65
Simplifying complex fractions uses all of the previous concepts about rational
expressions which we’ve covered in this chapter.
Example
Simplify.
x
x−
x+3
2
1+
x
x x x+3 x
x− ∗ −
x+3 1 x+3 x+3
2
= 1 x 2
creating common denominators
1+ ∗ +
x 1 x x
x(x + 3) − x
= xx +
+2
3

x
x2 + 3x − x x2 + 2x
= xx +
+3
2
x+3
= x + 2 dividing fractions
x x
x2 + 2x x x(x + 2) x
= ∗ = ∗
x+3 x+2 x+3 x+2
(x+2)
x
x
= ∗ factor and cancel to reduce to lowest terms
x+3 x+2

x2
=
x+3
Exercises 1.8
Simplify each complex fraction. Express your answer in lowest terms.
1 1 m

1 + 1+
1) y
2) yx y
x
3)
n
n2

x+ − 1−
2 x y m2
x−y
1 1
x 7 3

− − −
4) 1x y
1
5) yy x+y
x−y
6) a3 + 1 1 a
− + +
x2 y2 x x+y a a−1
1 1 1 4

x− − x+
2x + 1 2x − 2 x x+4
7) 2
8) 2 1
9) 4x + 4

1− − x−
2x + 1 x x−1 x+4
x+6
1 1
1

x− − 1−
10) x+2
4x + 15
11) x + 2 1x − 3 12) x+1
1

x− 1+ 2 1+
x+2 x −x−6 x−1
3
1 3
5

− y2 − 4
n+2−
13) a −
2
b a+b
4
14) 1 1
15) 1
n−2

+ − 1−
b−a b+a y+2 y−2 (n − 2)2
1 x−2 1 1

4+ 2+ −
x+1 1 − x2 2x − 1 2x + 1
16) 1
17) 3
18) 1

16 − 2− 4−
(x + 1)2 x+1 x2
1.9. RATIONAL EQUATIONS 67
1.9 Rational Equations
We can also use the skills we have covered in previous sections to solve equations
involving rational expressions. There are three main methods of solution that will
be explored in this section - multiplying on both sides to clear a denominator,
cross-multiplying, and making common denominators. Each of these techniques
is actually the same process, but approached from a slightly different perspective.
Clearing a denominator
Often, if the denominator is a single variable, it can be easy and striaghtforward
to multiply on both sides by the denominator to cancel it out.
Example
Solve for x.
4
x+ =7
x
If we multiply on both sides by x, it will clear the variable from the denominator:
4
x+ =7
x
4
x∗ x+ = (7) ∗ x Multiply on both sides by x
x
4
x∗x+x∗ = 7x Distribute the x
x
x2 + 4 = 7x
x2 − 7x + 4 = 0 Standard form
x ≈ 0.628, 6.372
If a problem is stated simply as the equality of two fractions, cross-multiplying

can be a useful method of solution.
Example
Solve for x.
x 7
=
2x + 3 x−4
x 7
=
2x + 3 x−4
x(x − 4) = 7(2x + 3)
x2 − 4x = 14x + 21
x2 − 18x − 21 = 0
x ≈ 19.100, −1.100
Cross-multiplying is really just a short-cut method of clearing out the denomina-

tors by multiplying on both sides by both denominators:
x 7
=
2x + 3 x−4
x 7
(x − 4)(2x + 3) ∗ = ∗ (x − 4)(2x + 3)
2x + 3 x−4
x 7
(x − 4)
(2x 3) ∗
+ = ∗
(x−4)(2x + 3)

2x+ 3

x −
4
x(x − 4) = 7(2x + 3)
Then the equation is ready to be solved as shown above - but by just cross-
mulitiplying, we skip directly to the solution portion of the problem.
Sometimes, it is helpful to create a common denominator in order to set up a

situation where cross-multiplying can be used.
Example
Solve for x.
1 4 3
+ =
x+6 x−2 x+1
1 4 3
+ =
x+6 x−2 x+1
1 x−2 4 x+6 3
∗ + ∗ =
x+6 x−2 x−2 x+6 x+1
1(x − 2) + 4(x + 6) 3
=
(x + 6)(x − 2) x+1
x − 2 + 4x + 24 3
=
(x + 6)(x − 2) x+1
5x + 22 3
=
(x + 6)(x − 2) x+1
(5x + 22)(x + 1) = 3(x + 6)(x − 2) = 3(x2 + 4x − 12)
5x2 + 27x + 22 = 3x2 + 12x − 36
2x2 + 15x + 58 = 0
x ≈ −3.75 ± 3.865i
Example
Solve for x.
2 x
+ =4
x − 2 2x − 1
2 x
+ =4
x − 2 2x − 1
2 2x − 1 x x−2
∗ + ∗ =4
x − 2 2x − 1 2x − 1 x − 2
2(2x − 1) + x(x − 2)
=4
(x − 2)(2x − 1)
4x − 2 + x2 − 2x
=4
(x − 2)(2x − 1)
x2 + 2x − 2 4
=
(x − 2)(2x − 1) 1
1(x2 − 2x − 2) = 4(x − 2)(2x − 1) = 4(2x2 − 5x + 2)
x2 − 2x − 2 = 8x2 − 20x + 8
0 = 7x2 − 18x + 10
x ≈ 1.760, 0.812
In some situations, we can create a single common denominator for every fraction
in the problem and then clear them all at once.
Example
Solve for x.
2 4 x
− = 2
x + 3 3x − 1 3x + 8x − 3
2 4 x
− = 2
x + 3 3x − 1 3x + 8x − 3
2 3x − 1 4 x+3 x
∗ − ∗ = 2
x + 3 3x − 1 3x − 1 x + 3 3x + 8x − 3
2(3x − 1) − 4(x + 3) x
=
(x + 3)(3x − 1) (x + 3)(3x − 1)
2(3x − 1) − 4(x + 3) = x
The missing step above is the clearing of both denominators:
2(3x − 1) − 4(x + 3) x
(x + 3)(3x − 1) ∗ = ∗ (x + 3)(3x − 1)
(x + 3)(3x − 1) (x + 3)(3x − 1)
2(3x − 1) − 4(x + 3) x
(x+3) (3x −
1) ∗ = ∗
(x+3) (3x −
1)

(x + 3) (3x − 1) (x + 3) (3x − 1)

2(3x − 1) − 4(x + 3) = x
As is true in the process of cross-multiplying, it isn’t necessary to actually cancel

out the denominators in completing the problem.
2 4 x
− = 2
x + 3 3x − 1 3x + 8x − 3
2 3x − 1 4 x+3 x
∗ − ∗ = 2
x + 3 3x − 1 3x − 1 x + 3 3x + 8x − 3
2(3x − 1) − 4(x + 3) x
=
(x + 3)(3x − 1) (x + 3)(3x − 1)
2(3x − 1) − 4(x + 3) = x
6x − 2 − 4x − 12 = x
2x − 14 = x
x = 14
Exercises 1.9
5 6
1) x+ = −6 2) x+ = −7
x x
5 7
3) y− =2 4) +1=a
y a
9 3 4 1
5) = 6) =
2y + 4 y 3n + 7 2
x 8 y−2 5
7) = 8) =
x+3 x+6 2 y−5
2 n x 2
9) = 10) =
n 5n + 12 4−x x
5x 1 a 1
11) = 12) =
14x + 3 x 8a + 3 3a
9 2 1 1 4 1
13) − = 14) + =
x−1 x+4 x+2 x−2 x+5 x−3
5 1 3 1 4 6
15) + = 16) − =
3x + 2 x − 1 x+2 y − 2 2y + 5 y−1
5 1 1 2
17) + =3 18) − =1
x+1 x+2 2x − 1 x + 7
6 1 10 3
19) − =3 20) + =2
y−4 y+2 a+1 a−2
3a a 2a
21) − =
a2 − 2a − 15 a + 3 a−5
u2 + 2 3u −2u − 1
22) 2
− =
u +u−2 u+2 u−1
4 2 5
23) + = 2
2x − 1 x + 3 2x + 5x − 3
5 2 2
24) − 2 =
x + 5 x + 2x − 15 x−3
5 3 4
25) − = 2
y − 2 2y − 1 2y − 5y + 2
x+2 x+4 2x + 1
26) + = 2
x−1 x x −x
x x+1 5
27) + 2 =
x + 2 x − 7x − 18 x−9
2a a 5
28) − = 2
a+7 a+3 a + 10a + 21
x−1 2x − 3 3
29) − = 2
2x + 1 x+3 2x + 7x + 3
y 6 y2 + 4
30) + = 2
y+4 y+1 y + 5y + 4
Addendum - Word Problems

Following are a selection of word problems - some from ancient times, some from
the Renaissance and Enlightenment, and some from the 19th and 20th century.
1) A teacher agreed to work 9 months for $562.50 and board. At the end of
the term, on account of two months absence caused by illness, he received only
$409.50 for his seven months work. If the teacher used all nine months of his
board during the term, what was his board per month? (American 1892)
2) A servant is promised $100 plus a cloak as wages for a year. After seven
months, he leaves and receives $20 plus the cloak. How much is the cloak worth?
(Clavius, German 1608)
1
3) The sales tax on garments is 20 of their value. A certain man buys 42 gar-
ments, paying in copper coins. Two garments and 10 copper coins are paid as
tax. What is the price of a garment, O learned one? (Ancient India)
4) Two wine merchants enter Paris, one of them with 64 casks of wine, the
other with 20 casks (all of the same value). Since they do not have enough money
to pay the customs duties, the first pays 5 casks of wine and 40 francs, and the
second pays 2 casks of wine and receives 40 francs change. What is the price of
each cask of wine and what is the duty on each cask? (Chuquet, French 1484)
5) One of two men had 12 fish and the other had 13 fish, and all of the fish
were of the same price. From the first man, a customs agent took away 1 fish and
12 denarii for payment. From the other man he took 2 fish and gave him back 7
denarii as change. Find the customs fee and the price of each fish.
(Fibonacci, Italian 1202)
6) Two traders transporting sheepskins approach their country’s border. The

first trader has 100 sheepskins and the border guard takes 10 sheepskins plus $25
as a tariff. The second trader has 42 sheepskins and for a tariff, the border guard
takes 7 sheepskins but returns $14 change. What is the tariff per sheepskin and
what is the value of each sheepskin?
7) Two people have a certain amount of money. The first says to the second,
“If you give me 5 denarii, I will have 7 times what you have left.” The second
says to the first, “If you give me 7 denarii, I will have 5 times what you have left.”
How much money does each have? Round to the nearest 10th.
(Leonardo, Italian c. 1500)
8) Two different scenarios from Ancient Greece:
Two friends were walking. One said to the other, “If I had 10 more coins, I would
have 3 times as much money as you.” The other said, “If I had 10 more coins, I
would have five times as much as you.” How many coins does each have?
Two friends were walking. One said to the other, “If you give me 10 of your coins,
I would have 3 times as much money as you.” The other said, “If you give me 10
of your coins, I would have five times as much as you.” How many coins does
each have?
9) Andy and Betty together have $6 less than Christine. If Betty gives $5 to
Andy, then Andy will have half as much as Christine. If, instead, Andy gives $5
to Betty, then Andy will have one-third as much as Betty. How much does each
person have to begin with?
10) Three friends (A, B and C) each have a certain amount of money. A says,
“I have as much as B plus one-third as much as C.” B says, “I have as much as C
plus one-third as much as A.” C says, “I have 10 more than one-third of B.” How
much does each person have? (Ancient Greece)
11) On a test, 39 more pupils passed than failed. On the next test, 7 who passed
the first test failed and one-third of those who failed the first test passed the sec-
ond. As a result, 31 more passed the second test than failed it. What was the
record of passing and failing on the first test?
12) At two stations, A and B, six miles apart on the railway, the prices of coal
are $20 per ton and $24 per ton respectively. The rates of cartage of coal are $2.00
per ton per mile from A and $3.00 per ton per mile from B. At a certain customer’s
home, on the railroad from A to B, the cost of coal is the same whether delivered
from A or B. Find the distance to this home from A.
13) There were three-fourths as many women as there were men on the train.
At the next station six men and eight women got off the train, and twelve men
and five women got on. There were then three-fifths as many women as men on
the train. How many men and how many women were originally on the train?
14) If a theater could put 5 more seats in a row, it would need 20 rows less,
but if each row had 3 fewer seats, it would take 20 rows more to seat the same
number. How many people will it seat?
15) An audience of 450 people is seated in rows, with the same number of peo-
ple in each row. It would take 5 rows less if 3 more people were seated in each
row. In how many rows are the people seated?
16) If evergreens are planted 4 feet closer together it will take 44 more trees for
a certain piece of road, but if they are planted 6 feet farther apart, it will take 44
fewer trees for the same length of road. How many miles is the piece of road?
(use 5280 feet = 1 mile)
17) A movie theater owner found that by raising the price of each ticket by
$1.00, 200 fewer people attened and she broke even, but that if she lowered the
price by $1.00 per person, 550 people attended and she increased her receipts by
$1000. What is the usual rate per person?
18) The brine pipes in an 84-foot width artificial hockey rink are equally spaced.
If the space between each pair of pipes were increased by 1 inch, then 84 fewer
lengths of pipe would be needed. What is the distance between the pipes now?
19) A living room shelf is 36 inches long and contains a certain number of
books of uniform width. If each book were one-half inch narrower, the shelf
would hold six more books. How many books of the wider variety does it hold?
20) I am a brazen lion; my spouts are my two eyes, my mouth and flat of my
right foot. My right eye fills a jar in two days, my left eye in three and my foot
in four. My mouth is capable of filling it in six hours. Tell me how long all four
together will take to fill it. (Ancient Greece)
21) A man wishes to have 500 rubii of grain ground. He goes to a mill that has
five stones. The first stone grinds 7 rubii of grain in an hour, the second grinds
5 rubii in an hour, the third 4 rubii in an hour, the fourth grinds 3 rubii per hour
and the fifth grinds 1 rubii per hour. In how long will the grain be ground and
how much is done by each stone? (Clavius, German 1583)
22) If two men and three boys can plow an acre in one-sixth of a day, how long
would it require three men and two boys to plow it?
(Edward Brooks, American 1873)
23) A cobbler can cut leather for ten pairs of shoes in one day. He can finish 5
pairs of shoes in one day. How many pairs of shoes can he both cut and finish in
one day? (Ancient Egypt)
24) Four waterspouts are filling a tank. Of the four spouts, one can fill the tank
in one day, the second takes two days, the third takes three days and the fourth
takes four days. How long will it take all four spouts working together to fill the
tank? (Ancient Greece)
25) If, in one day, a person can make 30 arrows or fletch [put feathers on] 20
arrows, how many arrows can this person both make and fletch in one day? (An-
cient China)
26) One military horse and one ordinary horse can pull a load of 40 dan. Two
ordinary horses and one inferior horse can pull the same load of 40 dan as can
three inferior horses and one military horse. How much can each horse pull in-
dividually? (Ancient China)
27) A barrel of water has several holes in it. The first hole empties the full bar-
rel in 3 days. The second hole empties the full barrel in 5 days. The third hole
empties the full barrel in 20 hours and the last hole empties the full barrel in 12
hours. With all the holes open, how long will it take to empty the barrel?
(Levi ben Gershon, French 1321)
28) A certain lion could eat a sheep in 4 hours; a leopard could do so in 5 hours;
and a bear in 6 hours. How many hours would it take for all three animals to de-
vour a sheep if it were thrown in among them? (Fibonacci, Italian 1202)
29) Two ships are some distance apart, which journey the first can complete in
5 days and the other in 7 days - it is sought in how many days they will meet if
they begin the journey at the same time. (Fibonacci, Italian 1202)
30) Sarah, alone, can paint the garage in 24 hours, her sister Jenny, alone, can
paint the same garage in 12 hours. With the aid of their mother, the three together
can paint the garage in 4 hours. How long would it take their mother, working
alone, to paint the garage?
31) It required 75 workers 38 days to build an embankment to be used for flood

control. Had 18 workers been removed to another job at the very start of opera-
tions, how much longer would it have taken to build the embankment?
32) Mark, alone, requires 6 hours to paint a fence; however, his younger brother,
who alone could do it in 9 hours, helps him. If they start work at 8:30am, at what
time should they finish the work?
33) A group decides to build a cabin together. The job can be done by 3 skilled
workers in 4 days or by 5 amateurs in 6 days. How long would it take if they all
work together?
34) If it requires 18 workers 50 days to build a piece of road, how many days
sooner would it be done if 7 more workers were hired at the beginning of opera-
tions?
35) A contractor estimated that a certain piece of work would be done by 9

carpenters in 8 hours or by 16 amateurs in 9 hours. The contractor wishes to get
the job done as quickly as possible and uses both professional carpenters and
amateurs on the same job. Four carpenters and 4 amateurs begin work at 6am.
Allowing 45 minuntes for lunch, at what time should they finish the job?
36) Mrs. Ellis alone can do a piece of work in 6 days. Her oldest daughter
takes 2 days longer; her youngest daughter takes twice as long as her mother.
How long will it take to complete the job if all three work together?
37) If 5 men and 2 boys work together, a piece of work can be completed in
one day and if 3 men and 6 boys work together, it can be completed in one day.
How long would it take a boy to do the work alone?
38) A coal company can fill a certain order from one mine in 3 weeks and from
a second mine in 5 weeks. How many weeks would be required to fill the order
if both mines were used?
39) If 25 skilled workers work for 8 days, they can complete the construction
of a concrete dam; 12 skilled workers and 15 untrained workers together can
complete the dam in 10 days. How long would it take an untrained worker alone
to complete the work on the dam?
40) A and B working togehter can complete a piece of work in 30 days. After
they have both worked 18 days, however, A leaves and B finishes the work alone
in 20 more days. Find the time in which each can do the work alone.
41) A dump cart can haul enough gravel to fill a pit in 6 days. A truck can do
the same work in 2 days. How long would it take two dump carts and one truck
working together to fill the pit?
Chapter 2
Polynomial and Rational Functions
This chapter will explore the solution of equations and inequalities involving
both polynomial and rational functions, primarily through the examination of
their graphical representations. We will also explore the use of polynomial long
division and synthetic division in breaking down polynomials into their prime
factors and the relationship between factors and roots.
2.1 Representing Intervals
Many of the problems in this chapter will have solutions that must be expressed
as an interval. This means a range of x values that will satisfy the original prob-
lem. In this section, we will introduce the translation of graphical intervals into
algebraic notation.
For example, in the diagram below, we would represent the interval shown on
the graph as x < −3.
−5 −4 −3 −2 −1 0 1 2 3 4 5
83
84 CHAPTER 2. POLYNOMIAL AND RATIONAL FUNCTIONS
In this diagram, we would represent the interval shown on the graph as

1 ≤ x < 7.
−1 0 1 2 3 4 5 6 7 8 9
In the next diagram, we have two separate intervals to represent - and we will
need two separate statements to represent them.
−10 −8 −6 −4 −2 0 2 4 6 8 10
These intervals would be represented as x < −4 OR x ≥ 6.
Sometimes students try to represent the intervals above as 6 ≤ x < −4, however,
this expression would represent a single interval where x is both less than −4
and, at the same time, greater than or equal to +6. This is simply not possible, and
would result in the empty set, which is the reason that the OR portion is needed
in the correct answer.
Example
Represent the intervals indicated on the graph below:
−20 −15 −10 −5 0 5 10 15 20 25 30 35
On this graph, there is one interval beginning at −5 and ending at +5 and an-
other beginning at 30 and continuing to infinity. Thus, these intervals would be
represented as −5 < x ≤ 5 OR x ≥ 30.
Students familiar with another form of interval notation may wish to represent
this interval as (−5, 5]∪[30, ∞). Both forms of notation accomplish the same goal.
2.1. REPRESENTING INTERVALS 85
Exercises 2.1
In each problem below, represent the intervals indicated on the graph.
1)
−7 −4 0 5
2)
−5 −1 8 12
3)
−1 2
4)
−4 6
5)
−8 −6 3
6)
−5 −2 7
7)
−10 −4 1 6
8)
−6 −2 2 15
9)
−12 −7 5
10)
−6 −2 3
11)
0 4
12)
0
2.1. REPRESENTING INTERVALS 87
13)
−5 −3 −1 4
14)
−2 5 7
2.2 Solution by Graphing
In previous courses the solution of linear equations is covered, generally by sepa-

rating the variables and constants on opposite sides of the equation to isolate the
variable. In the previous chapter we examined the solution of quadratic equa-
tions, in which the variable is isolated using the technique of completing the
square. The major difference between these methods of solution is that, in the
solving of quadratic equations, we must contend with several different powers
of the variable which makes it considerably more difficult to isolate the variable.
There are formulas like the quadratic formula available to solve cubic (x3 ) and
quartic (x4 ) equations, however these formulas are somewhat cumbersome and
archaic. The primary method for solving eqautions of degree higher than 2 is
solution by graphing or by algorithm.
Solution by algorithm is a very interesting process as there are many different

algorithms available. Which algorithm is most appropriate often depends on the
types of equations being solved and the technology available to solve them. Two
major types of algorithms that rely on the graphincal representation of an equa-
tion are called “Double False Position” and “The Newton-Raphson Method.”
Many commonly available pieces of techonolgy use one of these methods. Since
we have graphing calculators available to us, we will focus on solution by graph-
ing.
Whether using a TI (Texas Instruments) or Casio graphing calculator, or a soft-

ware based graphing ustility such as Graph or Desmos, the solution of these
equations focuses on finding the x-intercepts of the graph since this is where
the y value is 0. The specific processes for solving equations using each of these
different tools will be covered in class.
Example
Solve for x.
x3 − 3x2 = 2x − 7
The first step is to move all terms to one side of the equation and set them equal
to zero.
x3 − 3x2 − 2x + 7 = 0
2.2. SOLUTION BY GRAPHING 89
Then we graph the function and look for which x values will make the y value
equal to 0.
y = x3 − 3x2 − 2x + 7
6
−1 1 2 3
−2
−4
In this graph, we can see three roots, or x-intercepts, where the graph crosses the
x-axis. These are the x-values that make the y-value equal to 0. We can use the
available technology to find these x-values.
y = x3 − 3x2 − 2x + 7
6
2
x ≈ −1.491 x ≈ 2.834
−1 1 2 3
x ≈ 1.657
−2
−4
So, we can see that the solutions to the eqaution x3 − 3x2 − 2x + 7 = 0 are
x ≈ −1.491, 1.657, 2.834.
We can check these answers by plugging them back into the original equation.
(−1.491)3 − 3 ∗ (−1.491)2 − 2 ∗ (−1.491) + 7 = −3.3146 − 3 ∗ 2.223 + 2.982 + 7
= −3.3146 − 6.669 + 2.982 + 7
= −0.0016
The result is not exactly 0 since our answers have been rounded off to the 1000th
place. If we wanted greater accuracy, then we should include greater accuracy in
the value of our answers.
(1.657)3 − 3 ∗ (1.657)2 − 2 ∗ (1.657) + 7 = 4.5495 − 3 ∗ 2.7456 − 3.314 + 7
= 4.5495 − 8.2368 − 3.314 + 7
= −0.0013
and
(2.834)3 − 3 ∗ (2.834)2 − 2 ∗ (2.834) + 7 = 22.7614 − 3 ∗ 8.03155 − 5.668 + 7
= 22.7614 − 24.09465 − 5.668 + 7
= −0.00125
2.2. SOLUTION BY GRAPHING 91
Exercises 2.2
Solve for x.
1) 3x3 − 7x2 − x + 1 = 0 2) 24x4 + 5x2 − 13x + 3 = 0
3) 2x3 − 2x2 − 28x + 51 = 0 4) 2x3 + 5x2 − 15x + 7 = 0
5) x4 − 4x3 + x2 + 6x + 1 = 0 6) x4 + 2x3 + x2 − x − 6 = 0
7) x4 − 5x3 − 3x2 + 17x − 9 = 0 8) 2x3 − 5x − 3 = 0
9) x4 − 4x3 − 7x2 − 36x − 18 = 0 10) 6x3 − 25x2 + 21x + 10 = 0
11) 2x4 − 5x3 + x2 + 4x − 2 = 0 12) x4 + x3 − 5x2 + x − 4 = 0

2.3 Solution of Polynomial Inequalities by Graphing
In this section, we will combine the concepts of the previous two sections to solve
polynomial inequalities. In Section 2.2, we solved equations by graphing and
finding the x-values which made y = 0. In solving an inequality, we will be
concerned with finding the range of x values that make y either greater than or
less than 0, depending on the given problem.
Example
Solve the given inequality.
2x3 + 8x2 + 5x − 3 ≥ 0
First, we graph the function:
10
y = 2x3 + 8x2 + 5x − 3
5
−3 −2 −1 1
−5
Then we identify the intervals of x-values that make the y value greater than or
equal to zero, as indicated in the problem.
2.3. SOLUTION OF POLYNOMIAL INEQUALITIES BY GRAPHING 93
10
y≥0 5
A B y≥0
−3 −2 −1 C 1
−5
The indicated roots of the function (A, B and C) are the x-values that make y
equal to zero. These points divide the graph between the regions where y is
greater than zero and the regions where y is less than zero. The solution to the
given inequlaity 2x3 + 8x2 + 5x − 3 ≥ 0 are A≤ x ≤B OR x ≥C.
When we find the values of A, B and C: A= −3, B≈ −1.366 and C≈ 0.366, we can
complete the solution to the problem.
2x3 + 8x2 + 5x − 3 ≥ 0
−3 ≤ x ≤ −1.366 OR x ≥ 0.366
Example
x4 − 2x3 − 5x2 + 8x + 3 ≤ 0
First, we graph the function:

−2 −1 1 2
y = x4 − 2x3 − 5x2 + 8x + 3
−5
In this problem, we’re looking for the intervals of x values that make y less than
or equal to zero. First, we identify the roots of the function:
x ≈ −2.034 x ≈ 1.806
−3 −2 −1 1 2 3
x ≈ −0.320 x ≈ 2.549
−5
Next, we’ll identify the intervals where the y values are less than zero:
−2 −1 1 2
y≤0 y≤0
−5
So, the solution to the original inequality is:
x4 − 2x3 − 5x2 + 8x + 3 ≤ 0
−2.034 ≤ x ≤ −0.320 OR 1.806 ≤ x ≤ 2.549
In the next example we’ll be looking to identify both the intervals where y is
greater than zero, and the intervals where y is less than zero.
Example
Determine the interval(s) for which x3 + 5x2 + 5x + 1 ≥ 0
Determine the interval(s) for which x3 + 5x2 + 5x + 1 < 0
Once again, we’ll start by graphing the function to find the roots:
2
x = −1 x ≈ −0.268
−4 −3 −2 −1 1 2 3
x ≈ −3.732
−2
Now that we’ve indentified the roots, we can determine where the y values are
greater than zero and where they’re less than zero.
For y ≥ 0, we can see that this corresponds to: −3.732 ≤ x ≤ −1 OR x ≥ −0.268
For y < 0, we can see that this corresponds to: x < −3.732 OR −1 < x < −0.268
Exercises 2.3
1) Determine the interval(s) for which x3 − 4x2 + 2x + 3 ≥ 0
Determine the interval(s) for which x3 − 4x2 + 2x + 3 < 0
2) Determine the interval(s) for which 4x3 − 4x2 − 19x + 10 ≥ 0
Determine the interval(s) for which 4x3 − 4x2 − 19x + 10 < 0
3) Determine the interval(s) for which x3 − 2.5x2 − 7x − 1.5 ≥ 0
Determine the interval(s) for which x3 − 2.5x2 − 7x − 1.5 < 0
4) Determine the interval(s) for which x3 − 3.5x2 + 0.5x + 5 ≥ 0
Determine the interval(s) for which x3 − 3.5x2 + 0.5x + 5 < 0

5) Determine the interval(s) for which 6x4 − 13x3 + 2x2 − 4x + 15 ≥ 0
Determine the interval(s) for which 6x4 − 13x3 + 2x2 − 4x + 15 < 0
6) Determine the interval(s) for which x4 − x3 − x2 + 3x − 5 ≥ 0
Determine the interval(s) for which x4 − x3 − x2 + 3x − 5 < 0
7) Determine the interval(s) for which 3x4 + 3x3 − 14x2 − x + 3 ≥ 0
Determine the interval(s) for which 3x4 + 3x3 − 14x2 − x + 3 < 0
8) Determine the interval(s) for which 4x4 − 4x3 − 7x2 + 4x + 3 ≥ 0
Determine the interval(s) for which 4x4 − 4x3 − 7x2 + 4x + 3 < 0

Determine the interval(s) that satisfy each inequality.
9) x3 + x2 − 5x + 3 ≤ 0 10) x3 − 7x + 6 > 0
11) x3 − 13x + 12 > 0 12) x4 − 10x2 + 9 < 0
13) 6x4 − 9x2 − 4x + 12 ≥ 0 14) x4 − 5x3 + 20x − 16 > 0
15) x3 − 2x2 − 7x + 6 ≤ 0 16) x4 − 6x3 + 2x2 − 5x + 2 ≤ 0
17) 2x4 + 3x3 − 2x2 − 4x + 2 > 0 18) x5 + 5x4 − 4x3 + 3x2 − 2 ≤ 0

2.4 Solution of Rational Inequalities by Graphing
In the previous section, we saw how to solve polynomial inequalities by graph-

ing. In this section, we will use similar methods to solve rational inequalities.
Rational inequalities involve ratios of polynomials or fractions. Because these
types of problems involve fractions, the graphs of the functions that we work
with will have what are known as asymptotes. This word comes from a Greek
root having to do with two lines that come very close to each other but never
meet.
The vertical asymptotes of a graph will appear at places where the original ex-
pression has a zero denominator. This means that the function is not defined at
those x values and so, rather than having a y value at that point, the graph has an
asymptote.
Example
x+2
Below is a graph of the function y =
x−1
−8 −6 −4 −2 2 4 6 8
−5
Rather than having a y value at the point where x = 1, the dotted line indicates
the asymptote where the function is not defined. In the previous section, we were
interested in finding the roots of the function because these are the places where
y = 0, and can be the dividing points between where the y values are greater than
zero (y > 0) and the where the y values are less than zero (y < 0).
2.4. SOLUTION OF RATIONAL INEQUALITIES BY GRAPHING 101
The importance of the asymptotes in analyzing rational functions is that, like the
roots, these represent x values that can be the dividing points between where
y > 0 and where y < 0.
Example
2
>0
x−3
First we examine the graph:
−8 −6 −4 −2 2 4 6 8
−5
Notice that the asymptote for this graph occurs at the value x = 3, because this is
the x value that creates a zero denominator. Also notice that the y values switch
from being negative to being positive across the asymptote.
There are no roots for this function because there are no x values that make y = 0.
For a fraction to be zero, the numerator must equal zero. In this example the
numerator is 2 and no value of x will make it equal zero. Therefore the only
possible dividing point on the graph is x = 3, and the solution to the inequality
is x > 3.
Example
x−2
>0
x−3
−8 −6 −4 −2 2 4 6 8
−5
In this inequality, there is again an asymptote at x = 3, but there is also a root at

2−2 0
the value x = 2, because when x = 2. y = = = 0. So we have two
2−3 −1
dividing points to consider, x = 2 and x = 3. We can see from the graph that
y > 0 for x < 2 or x > 3, so that is the solution to the given inequality.
Example
x2 − 2
>0
x−3
In this problem, we have the same asymptote as the previous two problems: x =
3. However, in this inequality, there are two roots, because there are two x values
that make the numerator equal zero.
√
x2 − 2 = 0 means that x2 = 2 and x = ± 2 ≈ ±1.414
We can see these roots on the graph.

15
10
−8 −6 −4 −2 2 4 6 8
−5
In the graph above, we can see the asymptote at x = 3 and the two roots at
x ≈ 1.414, −1.414.
The x values that make y > 0 are −1.414 < x < 1.414 OR x > 3.
Example
x−2
>0
x2 − 3
−8 −6 −4 −2 2 4 6 8
−5
The roots for this function are the x values that make the numerator equal zero:
x − 2 = 0, therefore x = 2, and we can see this root on the graph.

The asymptotes for the function are the x values that make the denominator equal
zero:
√
x2 − 3 = 0 means that x2 = 3 and x = ± 3 ≈ ±1.732
−8 −6 −4 −2 2 4 6 8
−5
Therefore the solution for the given inequality is:
−1.732 < x < 1.732 OR x > 2
Example
5x + 1
<0
x2 + 3x − 4
−8 −6 −4 −2 2 4 6 8
−5
Roots
5x + 1 = 0
5x = −1
x = −0.2 = − 15
Asymptotes
x2 + 3x − 4 = 0
(x + 4)(x − 1) = 0
x = −4, 1
−8 −6 −4 −2 2 4 6 8
−5
If we combine the algebraic analysis above with what we see in the graph, then
we know that the dividing points important to the solution of this inequality are
at x = −4, −0.2, 1. The intervals where the y values are less than zero are x < −4
OR −0.2 < x < 1.
Example
x2 + 2x − 1
≤0
x2 + 7x + 5
−8 −6 −4 −2 2 4 6 8
−5
Roots
x2 + 2x − 1 = 0
x ≈ −2.414, 0.414
Asymptotes
x2 + 7x + 5 = 0
x ≈ −6.193, −0.807
We can see that the dividing points important to the solution of the inequality are
x ≈ −6.193, −2.414, −0.807, 0.414. The intervals where the y values are less than
or equal to zero are −6.193 ≤ x ≤ −2.414 OR −0.807 ≤ x ≤ 0.414.
Exercises 2.4
Solve each inequality.
x+4 2x + 3
1) >0 2) <0
x2 − 8x + 12 x2 − 2x − 35
x2 − 5x − 14 2x2 − x − 3
3) <0 4) >0
x2 + 3x − 10 x2 + 10x + 16
3x + 2 x2 + 2x + 5
5) <0 6) >0
x2 + x − 5 x2 − 3x − 7
x3 + 9 x3 + 9
7) >0 8) >0
x2 + x − 1 x2 + x + 1
Solve each inequality.
x2 − 2x − 9 x2 + 4x + 3
9) >0 10) <0
3x + 11 2x + 1
x2 + x − 5 x3 + 2
11) >0 12) >0
x2 − x − 6 x2 − 2
x2 + 2x − 7 2x − x2
13) <0 14) <0
x2 + 3x − 6 x2 − 4x + 6
x2 − 7 x−5
15) <0 16) >0
x2 + 5x − 1 3x2 − 2x − 3
2.5 Finding Factors from Roots
One method of solving equations involves finding the factors of the polynomial
expression in the equation and then setting each factor equal to zero.
x2 + 8x + 15 = 0
(x + 5)(x + 3) = 0
x+5=0 x+3=0
x = −5 x = −3
In this process, the reasoning is that if (x + 5) times (x + 3) equals zero, then one
of those expressions must be equal to zero. In setting them equal to zero, we find
the solutions of x = −5, −3. Plugging them back into the factored expression we
see the following:
(−5 + 5)(−5 + 3) = 0 ∗ −2 = 0
and
(−3 + 5)(−3 + 3) = 2 ∗ 0 = 0
This process works in reverse as well. In other words, if we know a root of the
function, we can find factors for the expression.
Example
Find a quadratic equation that has roots of −2 and +3.
x = −2 x=3
x+2=0 x−3=0
(x + 2)(x − 3) = 0
x2 − x + 6 = 0
2.5. FINDING FACTORS FROM ROOTS 109
Roots that are fractions are a little trickier, but really no more difficult:
Example
Find a quadratic equation that has roots of −5 and 23 .
2
x = −5 x= 3
x+5=0 3x = 2
x+5=0 3x − 2 = 0
(x + 5)(3x − 2) = 0
3x2 + 13x − 10 = 0
Exercises 2.5
Find a quadratic equation that has the indicated roots.
3
1) 4, −1 2) −2, 7 3) 2
,1
4) − 15 , 23 5) 1
3
,3 6) −4, 25
1
7) 2
, − 72 8) −1, 53 9) − 23 , −3
10) − 23 , − 34 11) − 52 , 3 12) −6, −2

2.6. POLYNOMIAL LONG DIVISION 111
2.6 Polynomial Long Division
Polynomial long division has many similarities to numerical long division, so it

is important that we understand how and why numerical long division works
the way it does before discussing polynomial long division. First the HOW?
Given the numerical problem 87, 462÷38, the first step is to determine the highest
place value in the answer.
2
38)87, 462
Often the first step in numerical long division is to say “Does 38 divide into 8?”
“No.” “Does 38 divide into 87?” “Yes.” This tells us that the first digit in the
answer will be over the 7 in 87,462, and consequently will be in the thousands
place. Once we know that the first digit in the answer will be in the thousands
place, the next question is “How many thousands?” We can determine that 2 ∗
38 = 76 but 3∗38 = 114 (too big), so we know that the first digit in the answer will
be 2. Then we subtract, include the 4 and examine what is left over to continue.
2
38)87, 462
−76
114
Here, we see that 114 ÷ 38 = 3, so we know that the next digit in the answer will
be 3.
23
38)87, 462
−76
114
−114
0006
After including the 6, we can see that 38 does not divide evenly into 6, so we put
a zero as the next digit in our answer and proceed:
2 30
38)87, 462
−76
114
−114
0006
0
00062
Now that we have included all the digits from our original number 87,462, the
last step is to divide 38 into 62. This goes one time with 24 left over.
2 301
38)87, 462
−76
114
−114
0006
−0
00062
−38
00024
So, now we have the solution to the original problem 87, 462 ÷ 38 = 2, 301 R24 or
24
2, 301 38 .
The WHY? of the long division algorithm is somewhat hidden by the HOW? In
the first step, we are determining which place value will hold the first digit of our
answer. When we determine that 38 does divide into 87, this indicates that the
first digit in our answer will be the thousands place. Dividing 38 into 87 tells us
how many thousands there will be. Then we subtract:
87, 462
−76, 000
11, 462
Now we need to determine how many times 38 will divide into 11,462. We de-
cided on 300 ∗ 38 = 11, 400, then we subtract to see how much is left over:
11, 462
−11, 400
00, 062
We can see that we won’t need any tens in our answer, and that 38 divides into 62
one time with 24 left over, thus the answer is 2 thousands, 3 hundreds, no tens, 1
and a remainder of 24. To check the answer, we multiply 38 ∗ 2301 and add 24:
2, 301
×38
18408
6903
87438
+24
87462
Polynomial long division works in much the same way that numerical long divi-
sion does. Given a problem A ÷ B, the goal is to find a quotient Q and remainder
R so that A = B ∗ Q + R.
Let’s look at this with the example 2x4 + 7x3 + 4x2 − 2x − 1 ÷ x2 + 3x + 1 or:
2x4 + 7x3 + 4x2 − 2x − 1

x2 + 3x + 1
So, we are looking to answer the question:
A = B ∗ Q +R
2x4 + 7x3 + 4x2 − 2x − 1 = (x2 + 3x + 1) ∗ (?+?+?)+?
If we want to mulitply x2 + 3x + 1 times something and end up with 2x4 + 7x3 +

4x2 − 2x − 1, then what we multiply by is going to have to start with 2x2 , because
x2 ∗ 2x2 = 2x4 .
Now we’re working with this:
A = B ∗ Q +R
2x4 + 7x3 + 4x2 − 2x − 1 = (x2 + 3x + 1) ∗ (2x2 +?+?)+?
But the 2x2 doesn’t just get multiplied by the x2 , it will also get multiplied by the
3x and the 1. So now we have:
A = B ∗ Q +R
2x4 + 7x3 + 4x2 − 2x − 1 = (x2 + 3x + 1) ∗ (2x2 +?+?)+?
= 2x4 + 6x3 + 2x2 +?????
The issue this raises is that the next multiplication (? ∗ x2 +? ∗ 3x+? ∗ 1) needs to
add only 1x3 to our answer, because we need 7x3 and we already have 6x3 from
the previous multiplication. That means we’re going to want to multiply next by
1x:
A = B ∗ Q +R
2x4 + 7x3 + 4x2 − 2x − 1 = (x2 + 3x + 1) ∗ (2x2 + 1x+?)+?
= 2x4 + 6x3 + 2x2
= 1x3 + 3x2 + x
= 2x4 + 7x3 + 5x2 + 1x+???
In the next round of mutiplication, we’re going to want to bring the 5x2 down to
4x2 , so we’ll need to multiply by −1.
A = B ∗ Q +R
2x4 + 7x3 + 4x2 − 2x − 1 = (x2 + 3x + 1) ∗ (2x2 + 1x − 1) + 0
= 2x4 + 6x3 + 2x2
= 1x3 + 3x2 + x
= −1x2 − 3x − 1
= 2x4 + 7x3 + 5x2 − 2x − 1
Now we also know that the remainder is zero, because x2 + 3x + 1 divides evenly
into 2x4 + 7x3 + 4x2 − 2x − 1 and so:
2x4 + 7x3 + 4x2 − 2x − 1 = (x2 + 3x + 1) ∗ (2x2 + 1x − 1)
This method makes the reasoning behind dividing polynomials somewhat more
apparent than the long division process, but it is more cumbersome. The way
that polynomial long division is usually approached is as follows:

x2 + 3x + 1 2x4 + 7x3 + 4x2 − 2x − 1
Then, just as we did in the other method, we question “What should we multiply
x2 by to get 2x4 ?” Answer: “2x2 ” This is the first term in our answer:
2x2

x2 + 3x + 1 2x4 + 7x3 + 4x2 − 2x − 1
Then we multiply 2x2 (x2 + 3x + 1) and change all the signs to see what we’ll be
left with:
2x2

x2 + 3x + 1 2x4 + 7x3 + 4x2 − 2x − 1
− 2x4 − 6x3 − 2x2
This indicates that we’ll have the 2x4 we’ll need in our answer, as well six of the
seven x3 ’s and two of the four x2 ’s. We’ll now need 1x3 next:
2x2

x2 + 3x + 1 2x4 + 7x3 + 4x2 − 2x − 1
− 2x4 − 6x3 − 2x2
x3 + 2x2 − 2x
This means we’ll need to multiply by 1x:
2x2 + x

x2 + 3x + 1 2x4 + 7x3 + 4x2 − 2x − 1
− 2x4 − 6x3 − 2x2
x3 + 2x2 − 2x
− x3 − 3x2 − x
− x2 − 3x − 1
Here, we still need to pick up a −1x2 , which means that our next multiplication
will be with −1:
2x2 + x − 1

x2 + 3x + 1 2x4 + 7x3 + 4x2 − 2x − 1
− 2x4 − 6x3 − 2x2
x3 + 2x2 − 2x
− x3 − 3x2 − x
− x2 − 3x − 1
x2 + 3x + 1
0
Because x2 + 3x + 1 divides evenly into 2x4 + 7x3 + 4x2 − 2x − 1 we have a zero

remainder. In the next example there will be a non-zero remainder:
Example
Divide:
3x4 − 8x3 + 19x2 − 15x + 10

x2 − x + 4
First, we set up the problem:

x2 − x + 4 3x4 − 8x3 + 19x2 − 15x + 10
Then, we question: “What do we need to multiply x2 by to get 3x4 ?” Answer:

“3x2 ”
3x2

x2 − x + 4 3x4 − 8x3 + 19x2 − 15x + 10
Then, we multiply, change signs (subtract) and combine like terms:
3x2

x2 − x + 4 3x4 − 8x3 + 19x2 − 15x + 10
− 3x4 + 3x3 − 12x2
− 5x3 + 7x2 − 15x
Now we’ll need to multiply by −5x, change signs and combine like terms:
3x2 − 5x

x2 − x + 4 3x4 − 8x3 + 19x2 − 15x + 10
− 3x4 + 3x3 − 12x2
− 5x3 + 7x2 − 15x
5x3 − 5x2 + 20x
2x2 + 5x + 10
We need 2x2 so we’ll need to multiply by 2, change signs and combine like terms:
3x2 − 5x + 2

x2 − x + 4 3x4 − 8x3 + 19x2 − 15x + 10
− 3x4 + 3x3 − 12x2
− 5x3 + 7x2 − 15x
5x3 − 5x2 + 20x
2x2 + 5x + 10
− 2x2 + 2x − 8
7x + 2
Because there is no positive power of x that we can multiply x2 by to get 7x, then
this is our remainder: 7x + 2.
So:
A = B ∗ Q + R
3x4 − 8x3 + 19x2 − 15x + 10 = (x2 − x + 4) ∗ (3x2 − 5x + 2) + (7x + 2)

Exercises 2.6
Find the quotient in each problem.
y 3 − 4y 2 + 6y − 4 x3 − 5x2 + x + 15
1) 2)
y−2 x−3
x3 − 4x2 − 3x − 10 2x3 − 3x2 + 7x − 3

3) 4)
x2 + x + 2 x2 − x + 3
x4 + 2x3 − x2 + x + 6 x4 + x3 + 5x2 + 3x + 6
5) 6)
x+2 x2 + x − 1
2z 3 + 5z + 8 x5 + 3x + 2
7) 8)
z+1 x3 + 2x + 1
x4 + 2x3 + 4x2 + 3x + 2 2x4 + 3x3 + 3x2 − 5x − 3

9) 10)
x2 + x + 2 2x2 − x − 1
2y 5 − 3y 4 − y 2 + y + 4 3x5 − 4x3 + 3x2 + 12x − 10

11) 12)
y2 + 1 x2 + 2x − 1
5x4 − 3x2 + 2 3y 3 − 4y 2 − 3
13) 14)
x2 − 3x + 5 y 2 + 5y + 2
2.7 Synthetic Division
The process for polynomial long division (like the process for numerical long
division) has been separated somewhat from its logical underpinnings for a more
efficient method to arrive at an answer. For particular types of polynomial long
division, we can even take this abstraction one step further. Synthetic Division is
a handy shortcut for polynomial long division problems in which we are dividing
by a linear polynomial. This means that the highest power of x we are dividing by
needs to be x1 . This limits the usefulness of Synthetic Division, but it will serve
us well for certain purposes. Let’s examine where the coefficients in our answer
come from when we divide by a linear polynomial:
2x3

x−5 2x4 − 6x3 − 23x2 + 16x − 5
− 2x4 + 10x3
4x3 − 23x2
Notice that the first coefficient in the answer is the same as the first coefficient in
the polynomial we’re dividing into. This is because we’re dividing by a polyno-
mial in the form 1x1 − a. This also makes the power of the first term in the answer
one less than the power of the polynomial we are dividing into. Let’s see where
the subsequent coefficients in the answer come from:
2x3 + 4x2

x−5 2x4 − 6x3 − 23x2 + 16x − 5
− 2x4 + 10x3
4x3 − 23x2
− 4x3 + 20x2
− 3x2 + 16x
The next coefficient in the answer (4) comes from the combination of the −6 and
the +10. The +10 came from multiplying the 2 in the answer by the 5 in the
divisor x − 5. The next coefficient in the answer will be −3, which comes from
multiplying the 4 (in the answer) by the 5 (in the divisor) and combining it with
the −23 in the polynomial we’re dividing into:
2.7. SYNTHETIC DIVISION 121
2x3 + 4x2 − 3x

x−5 2x4 − 6x3 − 23x2 + 16x − 5
− 2x4 + 10x3
4x3 − 23x2
− 4x3 + 20x2
− 3x2 + 16x
3x2 − 15x
The last part of our answer will come from multiplying the −3 in the answer
times the 5 in the divisor (making −15) and combining this with the +16 in the
polynomial we’re dividing into:
2x3 + 4x2 − 3x + 1

x−5 2x4 − 6x3 − 23x2 + 16x − 5
− 2x4 + 10x3
4x3 − 23x2
− 4x3 + 20x2
− 3x2 + 16x
3x2 − 15x
1x − 5
−x+5
0
The process of Synthetic Division uses these relationships as a shortcut to finding

the answer. The set-up for a Synthetic Division problem is shown below:
5 2 −6 −23 16 −5
2x4 − 6x3 − 23x2 + 16x − 5

This set-up allows us to complete the division problem .
x−5
5 2 −6 −23 16 −5
The first coefficient in the answer is the same as the first coefficient in the polyno-
mial we’re dividing into:
5 2 −6 −23 16 −5
↓
2
To get the next coefficient, we multiply the 2 by the 5 to get +10 and flll this in
under the −6:
5 2 −6 −23 16 −5
↓ +10
2
Then, −6 + 10 = +4, which is the next coefficient in the answer:
5 2 −6 −23 16 −5
↓ +10
2 4
Then, we continue this process, multiplying the 4 by the 5 to get 20 and combin-
ing this with the −23: −23 + 20 = −3:
5 2 −6 −23 16 −5
↓ +10 +20
2 4 −3
Next, multiply the −3 by the 5 and combine the resulting −15 with the 16:
5 2 −6 −23 16 −5
↓ +10 +20 −15
2 4 −3 1
In the last step, multiply the 1 times the 5 and combine the result with the −5 in
the problem to get zero:
5 2 −6 −23 16 −5
↓ +10 +20 −15 5
2 4 −3 1 0
This last coefficient represents the remainder - in this case 0. The other numerals
in the answer represent the coefficients for the powers of x in the answer. On the
far right is the remainder, then the constant (x0 ) term, then the linear (x1 ) term
and so on:
5 2 −6 −23 16 −5
↓ +10 +20 −15 5
2x3 4x2 −3x1 1x0 0
2x4 − 6x3 − 23x2 + 16x − 5

= 2x3 + 4x2 − 3x + 1
x−5
Let’s look at another example:
Example
Use Synthetic Division to divide:
3x3 + 5x2 − 9x + 9
x+3
Since Synthetic Division is set up to divide by x − a, if we’re dividing by x + 3

we’ll need to use a −3 in the Synthetic Division:
−3 3 5 −9 9
↓
3
Then, 3 ∗ −3 = −9:
−3 3 5 −9 9
↓ −9
3 −4
Next, −4 ∗ −3 = +12:
−3 3 5 −9 9
↓ −9 12
3 −4 3
And this example also has a zero remainder:
−3 3 5 −9 9
↓ −9 12 −9
3 −4 3 0
The answer here is 3x2 − 4x + 3:
3x3 + 5x2 − 9x + 9
= 3x2 − 4x + 3
x+3
and
3x3 + 5x2 − 9x + 9 = (x + 3)(3x2 − 4x + 3)
Let’s look at an example that is a little bit different.
Example
6x4 + x3 + 9x2 + x − 2
2x + 1
Synthetic Division is set up to handle problems in which we are dividing by

1x − a. Clearly, this is not the case in this example, however we can work around
this. Another way of looking at setting up the synthetic division is that we use
the number that is the solution to x − a = 0. When we divided by x − 5, we used
+5. When we were dividing by x + 3, we used −3. So, if we’re going to divide by
2x + 1, we’ll use − 12 in the Synthetic Division:
− 12 6 1 9 1 −2
↓
6
Then, we proceed as usual:
− 21 6 1 9 1 −2
↓ −3 1 −5 2
6 −2 10 −4 0
Notice that, again, we have a zero remainder. Also, notice that each coefficient
in our answer has a common factor of 2, which was the coefficient of the x in
2x + 1, which we orginally were going to divide by. What we’ve done here is not
division by 2x + 1, but division by x + 21 .
So, in the end, our work tells us that:
6x4 + x3 + 9x2 + x − 2
= 6x3 − 2x2 + 10x − 4
x + 12
and

4 3 2 1
6x + x + 9x + x − 2 = x + (6x3 − 2x2 + 10x − 4)
2
Notice that if we factor out the common factor of 2 from our answer, we can
multiply it back into the x + 21 and get an answer for our original problem:

4 3 2 1
6x + x + 9x + x − 2 = x+ 2(3x3 − x2 + 5x − 2)
2
= (2x + 1)(3x3 − x2 + 5x − 2)
This means that:
6x4 + x3 + 9x2 + x − 2
= 3x3 − x2 + 5x − 2
2x + 1
Another thing to understand about Synthetic Division is that if there is a missing

power of x, then you should include a zero as the coefficient of that power.
Example
x3 + 4x − 6
x−2
Since there is no x2 term in the polynomial we’re divding into, we’ll enter a zero
as the coefficient for that term:
2 1 0 4 −6
↓
1
And then proceed as usual:
2 1 0 4 −6
↓ 2 4 16
1 2 8 10
So the answer for this problem is x2 + 2x + 8 R : 10.

Exercises 2.7
Use synthetic division to find the quotient in each problem.
x3 − 8x2 + 5x + 50 x3 + 5x2 − x − 14
1) 2)
x−5 x+2
x3 + 12x2 + 34x − 7 x3 − 10x2 + 23x − 6

3) 4)
x+7 x−3
x4 − 15x2 + 10x + 24 x4 − 3x3 + 4x2 − 36

5) 6)
x+4 x−3
x4 − 2x3 − x + 10 x4 − 16x2 − 5x − 24
7) 8)
x−2 x+4
2x4 − x3 + 2x − 1 3x4 + x3 − 3x + 1
9) 10)
2x − 1 3x + 1
3x4 − 8x3 + 9x2 − 2x − 2 6x4 − 7x3 + 5x2 − 17x + 10

11) 12)
3x + 1 3x − 2
2x3 + 7x2 + 6x − 5 3x4 − x3 − 21x2 − 11x + 6

13) 14)
2x − 1 3x − 1
2.8 Roots and Factorization of Polynomials
In this section we will use some of the skills we have seen in previous sections
in order to find all the roots of a polynomial function (both real and complex)
and also factor the polynomial as the product of prime factors with integer coef-
ficients.
Example
Find all real and complex roots for the given equation. Express the given polyno-
mial as the product of prime factors with integer coefficients.
2x3 − 3x2 + 2x − 8 = 0
First we’ll graph the polynomial to see if we can find any real roots from the
graph:
20
10
x=2
−4 −2 2 4
−10
−20
We can see that there is a root at x = 2. This means that the polynomial will
have a factor of (x − 2). We can use Synthetic Division to find any other factors.
Because x = 2 is a root, we should get a zero remainder:
2 2 −3 2 −8
↓ 4 2 8
2 1 4 0
2.8. ROOTS AND FACTORIZATION OF POLYNOMIALS 129
So, now we know that 2x3 − 3x2 + 2x − 8 = (x − 2)(2x2 + x + 4). To finish the
problem, we can set each factor equal to zero and find the roots:
2x3 − 32 + 2x − 8 = 0
(x − 2)(2x2 + x + 4) = 0
x−2=0 2x2 + x + 4 = 0
x=2 x ≈ −0.25 ± 1.392i
Let’s look at an example that has more than one real root:
Example
3x4 + 5x3 − 45x2 + 19x − 30 = 0
graph:
400
200
x = −5 x=3
−8 −6 −4 −2 2 4 6 8
−200
−400
We can see roots at x = −5, 3, which means that (x+5) and (x−3) are both factors
of this polynomial. We’ll need to divide by both of these factors to break down
the polynomial. First, we divide by (x − 3):
3 3 5 −45 19 −30
↓ 9 42 −9 30
3 14 −3 10 0
And then by (x + 5):
3 3 5 −45 19 −30
↓ 9 42 −9 30
−5 3 14 −3 10 0
↓ −15 5 −10
3 −1 2 0
Now we know that 3x4 + 5x3 − 45x2 + 19x − 30 = (x + 5)(x − 3)(3x2 − x + 2) and
so, to finish the problem:
3x4 + 5x3 − 45x2 + 19x − 30 = 0
(x + 5)(x − 3)(3x2 − x + 2) = 0
x+5=0 x−3=0 3x2 − x + 2 = 0
1
x = −5 x=3 x≈ 6
± 0.799i
Next, let’s look at an example where there is a root that is not a whole number:
Example
3x3 + x2 + 17x + 28 = 0
graph:
40
20
x = −1.3
−4 −2 2 4
−20
−40
We can see in the graph that this polynomial has a root at x = − 43 . That means
that the polynomial must have a factor of 3x + 4. We can use Synthetic Division
to find the other factor for this polynomial. Because we know that x = − 34 is a
root, we should get a zero remainder:
− 34 3 1 17 28
↓ −4 4 −28
3 −3 21 0
Notice that, because the root we used was a fraction, there is a common factor of
3 in the answer to our Synthetic Division. We should factor this out to obtain the
answer:
(x + 43 )(3x2 − 3x + 21) = (3x + 4)(x2 − x + 7)
So, this means that:
3x3 + x2 + 17x + 28 = 0
(3x + 4)(x2 − x + 7) = 0
3x + 4 = 0 x2 − x + 7 = 0
4
x=− x ≈ 0.5 ± 2.598i
3
Exercises 2.8
Set #1
1) x4 − 3x3 + 5x2 − x − 10 = 0
2) 3x3 − 5x2 + 2x − 8 = 0
3) 2x4 − 5x3 + x2 + 4x − 4 = 0
4) x4 + x3 − 3x2 − 17x − 30 = 0
5) x4 − 9x3 + 21x2 + 21x − 130 = 0
6) x4 − 7x3 + 14x2 − 38x − 60 = 0
7) x5 − 9x4 + 31x3 − 49x2 + 36x − 10 = 0
8) x4 + 4x3 + 2x2 + 12x + 45 = 0

9) x4 − 6x3 + 12x2 − 10x + 3 = 0
10) x4 − 6x3 + 13x2 − 24x + 36 = 0
11) x5 − 3x4 + 12x3 − 28x2 + 27x − 9 = 0
12) x5 + 2x4 − 3x3 − 3x2 + 2x + 1 = 0
Set #2
13) 15x3 − 7x2 + 13x + 3 = 0
14) x4 − 5x3 + 3x2 − 11x − 20 = 0
15) 6x3 + 13x2 + 12x + 4 = 0
16) 6x3 − 5x2 + 5x − 2 = 0
17) 4x4 + 20x3 + 29x2 + 10x − 15 = 0

18) 3x4 − 4x3 + 10x2 + 12x − 5 = 0
19) 2x4 − 3x3 − 6x2 − 8x − 3 = 0
20) 12x4 − 53x3 − 31x2 − 19x − 5 = 0
21) 12x4 + 4x3 + x2 − 3x − 2 = 0
22) 3x4 + 13x3 − 26x − 40 = 0

Chapter 3
Exponents and Logarithms
In this chapter, we will examine concepts that are related to exponential, logarith-
mic and logistic relationships. In the first section, we will look at how to approach
these problems from a graphical perspective. In the subsequent sections, we will
examine the methods necessary to work with these problems algebraically.
3.1 Exponential and Logistic Applications
There are a variety of different types of mathematical relationships. The simplest

mathematical relationship is the additive relationship. This is a situation in which
the value of one quantity is always a certain amount more (or less) than another
quantity. A good example of an additive relationship is an age relationship. In
an age relationship, the age of the older person is always the same amount more
than the age of the younger person. If the older person is five years older, then
the age of the older person (y) will always be equal to the age of the younger
person (x) plus five: y = x + 5.
Another type of additive relationship is seen where two quantities add up to a

constant value. Let’s say there is a board whose length is 20 inches. If we cut
it into two pieces, with one piece being 6 inches, then the other piece will be 14
inches. If one piece is 9 inches, then the other will be 11 inches. If one piece is x
inches, then the other piece (y) will be 20 − x: y = 20 − x or x + y = 20.
135
136 CHAPTER 3. EXPONENTS AND LOGARITHMS
The next type of mathematical relationshp is a multiplicative relationship. This

represents a situation in which one quantity is always a multiple of the other
quantity. This is commonly seen in proportional relationships. If a recipe for a
cake calls for 2 cups of flour, then, if we want to make 3 cakes, we’ll need 6 cups
of flour. The amount of flour (y) is always two times the number of cakes we
want to make (x): y = 2x.
If a recipe for a batch of cookies (with 20 cookies per batch) calls for 1.5 cups
of sugar, then three batches would require 4.5 cups of sugar. The amount of
sugar required (y) is always the number of batches (x) times 1.5: y = 1.5x. If we
wanted to represent this relationship based on the number of cookies instead of
the number of batches, we would need to adjust the formula. Given that there are
20 cookies per batch, we could adjust our formula so that we first calculate the
number of batches from the number of cookies and then multiply by 1.5. If the
x 3
number of cookies is x and the amount of sugar is y, then y = 1.5 ∗ 20 or y = 40 x.
The next type of mathematical relationship is the polynomial relationship. In this

type of relationship, one quantity is related to a power of another quantity. A
good example of this type of relationship involves gravity. As Galileo discov-
ered in the 16th century, the distance that an object falls after it is dropped is not
proportional to the time that it has been falling. Rather, it is proportional to the
square of the time. The table below shows this type of relationship.
t d
1 16
2 64
3 144
4 256
After one second, it looks like the distance will always be sixteen times the time
the object has been falling. However, after two seconds, we can see that this
relationship no longer is true. That’s because this relationship is a polynomial
relationship in which the distance an object has fallen (d) is proportional to the
square of the time it has been falling (t): d = 16t2 .
3.1. EXPONENTIAL AND LOGISTIC APPLICATIONS 137
Exponential Relationships
The next type of relationship is the focus of this chapter - the exponential relation-
ship. In this situation, the rate of change of a quantity is proportional to the size
of that quantity. This relationship can be explored in more depth in an integral
calculus course, but we will discuss the basics here.
In a linear or proportional relationship, the slope, or rate of change, is constant.

For example, in the equation y = 3x + 1, the slope is always three, no matter what
the values of x and y are. In an exponential relationship, the rate of change (also
called ”y prime” or y 0 ) is proportional to the value of y. In this case, we say that
y 0 = k ∗ y.
This is what is known as a differential equation. This is an equation in which the

variable and its rate of change are related. Through the processes of differential
and integral calculus, we can solve the equation above y 0 = k ∗ y as:
y = Aekt
In the equation above, A is the value of y at time t = 0, k is a constant that

determines how fast the quantity y increases or decreases and t plays the role
of the independent variable (as x often does) and represents the time that has
passed. If k is positive, then the quantity y is growing because its rate of change
is positive. If k is negative, then the quantity y is decreasing because the rate of
change is negative.
The quantity represented by e in the above equation is a mathematical constant

(like π) that is often used to represent exponential relationships. The best way to
understand the value of e and what it represents is directly related to fundamen-
tal questions from differential and integral calculus.
Differential Calculus is concerned primarily with the question of slopes. We dis-

cussed earlier that a linear relationship has a constant slope. Polynomial and ex-
ponential relationships have slopes that depend on the value of x and/or y. This
is what makes them curves rather than lines. If we consider the slopes of some
different exponential relationships, we can see one aspect of where the value for
e comes from.
Consider the graphs of the following relationships:
y = 1.5x
y = 2x
y = 3x
y = 10x
Let’s look at the graphs for these functions:
20 20
15 15
10 y = 1.5x 10 y = 2x
5 5
−4 −2 2 4 −4 −2 2 4
−5 −5
20 20
15 15
10 y = 3x 10 y = 10x
5 5
−4 −2 2 4 −4 −2 2 4
−5 −5
We can see that these graphs demonstrate slightly different behavior and differ-
ent x and y values. One thing that they all have in common is that they all pass
through the point (0, 1) on the graph. This is because 1.50 = 1, 20 = 1, 30 = 1 and
100 = 1. Therefore the point where x is 0 and y is 1 is on all four of the graphs.
Although all four of the graphs pass through the point (0, 1), they each do this
in a different way. Let’s look at the slope of a line tangent to each curve at the
point (0, 1). This is the straight line that touches the curve at the point (0, 1), but
nowhere else:
20 20
15 15
10 y = 1.5x 10 y = 2x
5 5
−4 −2 2 4 −4 −2 2 4
−5 −5
20 20
15 15
10 y = 3x 10 y = 10x
5 5
−4 −2 2 4 −4 −2 2 4
−5 −5
We can see that the slopes of these tangent lines are all different. In the case of
y = 2x , the slope of the tangent line at (0, 1) is about 0.7, while for the graph of
y = 3x , the slope of the tangent line at (0, 1) is about 1.1.
As mathematicians examined these graphs during the 17th and 18th centuries,
they began to question what the value of the base ”b” should be in the equation
y = bx so that the slope of the tangent line at the point (0, 1) would be equal to
exactly 1. The answer was e ≈ 2.71828.
Another way to derive the value of e uses Integral Calculus. Integral Calculus is
often concerned with finding the area under a curve. This process can then be
generalized and used to make many other types of calculations that are similar
to finding area.
1
Consider the graph of the curve y = :
x
1
1 y= x
−1 1 2 3 4 5
−1
We can delineate borders on the x values and determine the area of the resulting
region using the techniques of calculus:
1
1 y= x 1
1
0.8 0.8 y= x
0.6 0.6
0.4 A ≈ 0.693 0.4
0.2 0.2 A ≈ 1.0986
0.6 0.8 1 1.2 1.4 1.6 1.8 2 1 1.5 2 2.5 3

−0.2 −0.2
These values for the area under the curve are actually the same values as those
for the slope of the tangent line in the previous graphs. If you ask the question,
”Where should you draw the second vertical line so that the area under the curve
is equal to exactly 1?” then just like the slope question, the answer is e ≈ 2.71828.
This is how the value of e was determined and why it is used to represent these
exponential relationships.
Logistic Relationships
Let’s consider the graph of y = 2x :
20
15
10
−4 −2 2 4
−5
If we extend the x-axis out further past x = 4, we would see that the y values for
this relationship will grow very quickly, as they continue doubling.
1,000
800
600
400
200
−4 −2 2 4 6 8 10
Some phenomena in the natural world exhibit behavior similar to the growth
of this function. However, in the natural world, few, if any, things can grow
unconstrained. Most growth of any kind is limited by the resources that fuel
the growth. Populations often grow exponentially for a period of time, however,
populations are dependent on natural resources to continue growing. As a result,
the simple exponential function is only useful for modeling real-world behavior
if the x-values are limited.
It was this problem with the simple exponential function that led French mathe-
matician Pierre Verhulst to slightly adjust the differential equation that gives rise
to the exponential function to make it more realistic.
The original differential equation said:
y0 = k ∗ y
This says that the rate of growth of y is always directly proportional to the value
of y. In other words the larger a population gets, the faster it will grow - forever.
Verhulst changed this to say:
y
y 0 = k ∗ y(1 − N
)
This is the defining relationship for the Logistic function. Notice that when values
of y are small, this is essentially the same as the simple exponential. If y is small,
then the (1− Ny ) term will be very close to 1 and will produce behavior very much
like the simple exponential.
The N in the formula is a theoretical “maximum population.” As the value of y

approaches this maximum value, Ny will approach 1 and (1 − Ny ) will get smaller
and smaller. As it gets smaller, the factor of (1 − Ny ) will slow down the growth
of the function to model the pressure that is put on the resources that are driving
the growth.
The solution of the Logistic equation is quite complicated and results in a stan-
dard form of:
N
y=
1 + be−kt
The graph for a sample logistic relationship is shown below:
0.8
0.6
1
0.4 y=
1 + e−0.1t
0.2
−40 −20 20 40
The “lazy-s” shape is characteristic of the logistic function. In the early stages,
the relationship shows growth very similar to the simple exponential function,
but, as the function grows larger, the growth decreases and the function values
stabilize. The maximum y value of N is always the horizontal asymptote for the
logistic function.
Negative Exponential Relationships

The Logistic function is very useful for modeling phenomena from the natural
world. Although the simple exponential function is somewhat limited in model-
ing natural phenomena, the negative exponential is quite useful. Looking back to
the graph of y = 2x :
20
15 y = 2x
10
−4 −2 2 4
−5
If we turn the graph around by changing x to −x in the formula, then we will be

working with the decaying tail of the graph:
20
15
10 y = 2−x
−4 −2 2 4
−5
Let’s zoom in on the portion of the graph in the Quadrant I:
1.5
1
y = 2−x
0.5
−1 1 2 3 4 5
−0.5
The behavior shown in the graph is quite useful for modeling radioactive decay,
processes of heating and cooling, and assimilation of medication in the blood-
stream. Let’s look at an example.
Example
If a patient is injected with 150 micrograms of medication, the amount of medi-

cation still in the bloodstream after t hours is given by the function:
A(t) = 150e−0.12t
a) Find the amount of medication in the bloodstream after 3 hours.
Round your answer to the nearest 100th.
b) How long will it take for the amount of medication to reach 60 micrograms?
Round your answer to the nearest 10th of an hour.
In this section we will focus on solving these problems using the graphing cal-
culator. In later sections, we will cover processes that can be used to solve these
problems algebraically. It is helpful to understand both methods of solution.
First, let’s graph the function given in the problem:
150
100
50
y = 150e−0.12x
5 10 15 20
We can directly calculate the answer for part (a) by plugging the value of 3 for t
or we can use the table in the graphing calculator to find this value:
x y
0 150
1 133.04
2 117.99
3 104.65
We can see that after 3 hours there are approximately 104.65 micrograms of med-
ication in the patient’s bloodstream.
To answer part (b) graphically, we’ll graph the original function along with the
horizontal line y = 60. When we find the intersection of these two graphs, we’ll
know how long it takes for there to be 60 micrograms of medication in the pa-
tient’s bloodstream:
150
100
(7.636, 60)
50
5 10 15 20
Here, we can see that it would take about 7.6 hours (or about 7 hours 36 minutes),
for there to be 60 micrograms of medication in the patient’s bloodstream.
Example
The deer population on a nature preserve can be modeled using the equation:
8000
y=
1 + 9e−0.2t
y indicates the number of deer living in the nature preserve and t represents the
number of years that have passed since the initial population of deer were estab-
lished there.
a) How many deer were in the initial population?
b) What is the deer population after 10 years?
Round your answer to the nearest whole number.
c) How long does it take for the population to reach 5,000?
Round your answer to the nearest 10th of a year.
Parts (a) and (b) can both be answered from a table of values for the function.
Before we look at the table, let’s consider the question in Part (a). The problem is
asking what the initial population of deer was. The means that we’re looking to
find out what y is when t = 0.
Let’s see what happens when we plug zero in for t in the formula:
8000 8000
y= −0.2∗0
=
1 + 9e 1 + 9e0
8000 8000
= =
1+9∗1 10
= 800
Let’s look at the table of values:
t y
0 800
5 1855.8
10 3606.8
To the nearest whole number, the deer population after 10 years is 3,607.
In Part (c), we’ll need to graph a horizontal line at y = 5, 000 and find the inter-
section.
8,000
6,000
(13.54, 5000)
4,000
2,000
5 10 15 20 25 30
To the nearest 10th, it takes about 13.5 years for the deer population to reach
5,000.
Exercises 3.1
1) Medication in the bloodstream
The function A(t) = 200e−0.014t gives the amount of medication, in milligrams, in

a patient’s bloodstream t minutes after the medication has been injected.
a) Find the amount of medication in the bloodstream after 45 minutes.
Round your answer to the nearest milligram.
b) How long will it take for the amount of medication to reach 50 milligrams?
Round your answer to the nearest minute.
2) Medication in the bloodstream
The function D(t) = 50e−0.2t gives the amount of medication, in milligrams, in a

patient’s bloodstream t hours after the medication has been injected.
a) Find the amount of medication in the bloodstream after 3 hours.
Round your answer to the nearest milligram.
b) How long will it take for the amount of medication to reach 10 milligrams?
Round your answer to the nearest minute.

3) Fish Population
The number of bass in a lake can be modeled using the given equation:
3600
P (t) =
1 + 7e−0.05t
where t is the number of months that have passed since the lake was stocked with
bass. In each question below, round your answer to the nearest whole number.
a) How many bass were in the lake immediately after it was stocked?
b) How many bass were in the lake 1 year after it was stocked?
4) Bird Population
The population of a certain species of bird is limited by the type of habitat re-
quired for nesting. The population can be modeled using the following equation:
5600
P (t) =
0.5 + 27.5e−0.044t
where t is the number of years. In each question below, round your answer to the
nearest whole number.
a) Find the initial bird population.
b) What is the population 100 years later?

5) A Temperature Model
A cup of coffee is heated to 180◦ F and placed in a room that maintains a temper-
ature of 65◦ F . The temperature of the coffee after t minutes have passed is given
by:
F (t) = 65 + 115e−0.042t
a) Find the temperature of the coffee 10 minutes after it is placed in the room.
Round your answer to the nearest degree.
b) When will the temperature of the coffee be 100◦ F ?
Round your answer to the nearest tenth of a minute.
6) A Temperature Model
Soup at a temperature of 170◦ F is poured into a bowl in a room that maintains

a constant temperature. The temperature of the soup decreases according to the
model given by:
F (t) = 75 + 95e−0.12t
where t is the number of minutes that have passed since the soup was poured.
a) What is the temperature of the soup after 2 minutes?
Round your answer to the nearest 10th of a degree.
b) A certain customer prefers that the soup be cooled to 110◦ F .
How long will this take?
Round your answer to the nearest 10th of a minute.

7) Radioactive Decay
A radioactive substance decays in such a way that the amount of mass remaining
after t days is given by the equation:
m(t) = 13e−0.015t
where the amount is measured in kilograms.
a) Find the mass at time t = 0.
b) How much of the mass remains after 45 days?
c) How long does it take for there to be 5 kg. left?
8) Radioactive Decay
Radioactive iodine is used by doctors as a tracer in diagnosing certain thyroid

gland disorders. This type of iodine decays in such a way that the mass remaining
after t days is given by:
m(t) = 6e−0.087t
where the amount is measured in grams.
a) Find the mass at time t = 0.
b) How much remains after 20 days?
c) How long does it take for there to be 2 grams left?
Round your answer to the nearest tenth of a day.

9) Fish population
The function:
12
P (t) =
1 + e−t
gives the size of a fish population in thousands at time t, measured in years.
a) Find the initial population of fish at time t = 0.
Find the population of fish after 2 years, time t = 2 .
b) How long will it take for the population to be 10,000?
What appears to be the maximum population for this particular model?
10) Fish population
The function:
70
P (t) =
1 + 2e−t
Find the population of fish after 1.5 years, t = 1.5.
What appears to be the maximum population for this particular model?

11) Fish population
The function:
20
P (t) =
1 − 0.4e−0.35t
Find the population of fish after 5 years, time t = 5.
What happens to this population over time?
12) Fish population
The function:
10
P (t) =
1 − 0.73e−0.08t
Find the population of fish after 5 years, time t = 5.
What happens to this population over time?

13) Continuous Mixing
A 100 gallon tank of pure water has salt water with 0.5 lb per gallon added to
it at a rate of 2 gallons per minute. The brine solution is mixed thoroughly and
drained at rate of 2 gallons per minute.
The equation for how many pounds of salt are in the tank at time t is given by:
y = 50 − 50e−0.02t
where y is measured in pounds and t is measured in minutes.
a) How many pounds of salt are in the tank after 20 minutes?
b) How long does it take for there to be 30 lbs of salt in the tank?
A 20 gallon tank of pure water has salt water with 0.4 lb per gallon of salt added
to it at a rate of 3 gallons per minute. The brine solution is mixed thoroughly and
drained at rate of 3 gallons per minute.
The equation for how many pounds of salt are in the tank at time t is given by:
y = 8 − 8e−0.15t
where y is measured in pounds and t is measured in minutes.
a) How many pounds of salt are in the tank after 15 minutes?

A tank contains 200 gallons of a 2% solution of HCl. A 5% solution of HCl is

added at 5 gallons per minute. The well mixed solution is being drained at 5
gallons per minute.
The amount of HCl in the tank at any given time t in minutes is:
y = 10 − 6e−0.025t
a) How many gallons of HCl are in the tank at t = 25 minutes?
b) When does the concentration of HCl in the solution reach 4%?
A tank contains 100 gallons of salt water which contains a total of 25 lbs of salt.
Salt water containing 0.4 lbs per gallon is added to the tank at a rate of 5 gallons
per minute and the well-mixed solution is drained at the same rate.
The amount of salt in pounds in the tank at any given time t in minutes is:
y = 40 − 15e−0.05t
a) How many pounds of salt are in the tank at t = 20 minutes?
3.2. LOGARITHMIC NOTATION 157
3.2 Logarithmic Notation
A Logarithm is an exponent. In the early 1600’s, the Scottish mathematician John

Napier devised a method of expressing numbers in terms of their powers of ten
in order simplify calculation. Since the advent of digital calculators, the methods
of calculation using logarithms have become obsolete, however the concept of
logarithms continues to be used in many area of mathematics.
The fundamental idea of logarithmic notation is that it is simply a restatement of

an exponential relationship. The definition of a logarithm says:
logb N = x → bx = N
The notation above would be read as ”log to the base b of N equals x means that
b to the x power equals N .” In this section we will focus mainly on becoming
familiar with this notation. In later sections, we will learn to use this process to
solve equations.
Example
Express the given statement using exponential notation:
log2 32 = 5
If log2 32 = 5, then 25 = 32.
Example
log7 4 ≈ 0.7124
If log7 4 ≈ 0.7124, then 70.7124 ≈ 4
If the logarithm notation appears without a base, it is usually assumed that the
base should be 10.
Example
log 100 = 2
If log 100 = 2, then 102 = 100
The notation ln N = x is typically used to indicate a logrithm to the base e. This

means that:
ln N = x → ex = N
Example
ln 15 ≈ 2.708
If ln 15 ≈ 2.708, then e2.708 ≈ 15
In some cases, we would want to change an exponential statement into a loga-

rithmic statement.
Exmple
Express the given statement using logarithmic notation:
124 = 20, 736
If 124 = 20, 736 then log12 20, 736 = 4

3.2. LOGARITHMIC NOTATION 159
Example
102.5 ≈ 316.23
If 102.5 ≈ 316.23 then log 316.23 ≈ 2.5
Example
e6 ≈ 403.4
If e6 ≈ 403.4, then ln 403.4 ≈ 6

Exercises 3.2
Rewrite each of the following using exponential notation.
1) t = log5 9 2) h = log7 10 3) log5 25 = 2
4) log6 6 = 1 5) log 0.1 = −1 6) log 0.01 = −2
7) log 7 ≈ 0.845 8) log 3 ≈ 0.4771 9) log2 35 ≈ 5.13
10) log12 50 ≈ 1.5743 11) ln 0.25 ≈ −1.3863 12) ln 0.989 ≈ −0.0111
Rewrite each of the following using logarithmic notation.
1
13) 102 = 100 14) 104 = 10, 000 15) 4−5 =
1024
1 3 1
16) 5−3 = 17) 16 4 = 8 18) 83 = 2
125
19) 101.3 ≈ 20 20) 100.301 = 2 21) e3 ≈ 20.0855
22) e2 ≈ 7.3891 23) e−4 ≈ 0.0183 24) e−2 ≈ 0.1353

3.3. SOLVING EXPONENTIAL EQUATIONS 161
3.3 Solving Exponential Equations
Because of the fact that logarithms are exponents, the rules for working with
logarithms are similar to those that govern exponential expressions. One very
helpful rule of equality for working with logarithms is related to the exponential
rule for raising a power to a power. We recall one of the rules of exponents as:
(bx )y = bx∗y
in other words
(52 )4 = (52 )(52 )(52 )(52 ) = 52∗4 = 58
In logarithmic notation, this rule works out as:
logb M p = p ∗ logb M
The reason for this comes from the rule for exponents. Lets’ say that logb M = x.
Then:
logb M = x
this means that
bx = M
and
(bx )p = (M )p
so
bp∗x = M p
Now, we come back to the question of logb M p =?. This expression (logb M p ) is
asking the question ”What power do we raise b to in order to get an answer of
M p ? The result on the previous page shows that:
bpx = M p
This means that we must raise b to the px power to get an answer of M p . Remem-
ber that x = logb M . This means that:
bpx = M p
so
logb M p = px = p ∗ logb M
This statement of equality is useful if we are trying to solve equations in which

the variable is an exponent.
Example
Solve for x.
4x = 53
We start by taking a logarithm on both of the equation. Just as we can add to both
sides of an equation, or multiply on both sides of an equation, or raise both sides
of an equation to a power, we can also take the logarithm of both sides. So long
as two quantities are equal, then their logarithms will also be equal.
4x = 53
log 4x = log 53
x log 4 = log 53
log 53
x= ≈ 2.864
log 4
Since the log base 10 and log base e are both programmed into most calculators,
these are the most commonly used bases for logarithms.
Example
Solve for x.
52x+3 = 17
We start this problem in the same fashion, but this time we will use a logarithm
to the base e:
52x+3 = 17
ln 52x+3 = ln 17
(2x + 3) ln 5 = ln 17
There are several possibilities for finishing the problem from this point. We will
focus on two of them that are the most useful for solving more complex problems.
First, we will distribute the ln 5 into the parentheses and then get the x by itself.
(2x + 3) ln 5 = ln 17
x ∗ 2 ln 5 + 3 ln 5 = ln 17
−3 ln 5 = −3 ln 5
x ∗ 2 ln 5 = ln 17 − 3 ln 5
ln 17 − 3 ln 5
x= ≈ −0.620
2 ln 5
And we can check the answer by plugging it back in:
52∗(−0.0620)+3 ≈ 51.760 ≈ 16.9897 ≈ 17

We can also approximate the logarithms in the problem and solve for an approx-
imate answer:
(2x + 3) ln 5 = ln 17
x ∗ 2 ln 5 + 3 ln 5 = ln 17
3.2189x + 4.8283 ≈ 2.8332
−4.8283 ≈ −4.8283
3.2189x ≈ −1.9951
x ≈ −0.620
If you use the method of approximating, it’s important to make a good approxi-
mation. At least 4-5 decimal places are necessary for an accurate answer.
Let’s look at an example that has variables on both sides of the equation:
Example
Solve for x.
43x = 92x−1
We’ll use log base 10 in this problem.
43x = 92x−1
log 43x = log 92x−1
3x ∗ log 4 = (2x − 1) log 9
x ∗ 3 log 4 = x ∗ 2 log 9 − log 9

If we collect like terms, we’ll end up with:
x ∗ 3 log 4 = x ∗ 2 log 9 − log 9
log 9 = x ∗ 2 log 9 − x ∗ 3 log 4
At this point, if we want to get the x by itself, we need to factor out the x on the
right-hand side:
log 9 = x ∗ 2 log 9 − x ∗ 3 log 4
log 9 = x(2 log 9 − 3 log 4)
Then divide on both sides by the coefficient in parentheses:
((
log 9 (2(
x( log
((9−
((3 log
(
4)
= ((
2 log 9 − 3 log 4 ( ((9 − 3 log 4
2 log ( ( (
log 9
=x
2 log 9 − 3 log 4
9.327 ≈ x
Again, we can check our answer by plugging it back into the equation:
43∗9.327 ≈ 427.981 ≈ 7.0184 ∗ 1016
92∗9.327−1 ≈ 917.654 ≈ 7.0177 ∗ 1016

We could also have solved this equation by approximating the logarithms in the
beginning.
43x = 92x−1
log 43x = log 92x−1
3x ∗ log 4 = (2x − 1) log 9
3x(0.60206) ≈ (2x − 1)0.95424
1.80618x ≈ 1.9085x − 0.95424
0.95424 ≈ 0.10232x
9.326 ≈ x
This answer is less accurate than the other approximation (9.326036 vs. 9.327424).
The accuracy of an answer depends upon the original approximations for the
logarithms.
Exercises 3.3
Solve for the indicated variable.
1) 2x = 5 2) 2x = 9 3) 3x = 7
4) 3x = 20 5) 2x+1 = 6 6) 7x+1 = 41
7) 5x+1 = 36 8) 8x−2 = 6 9) 42x+3 = 50
10) 4x+2 = 5x 11) 52x+1 = 9 12) 6x+4 = 10x
13) 7y+1 = 3y 14) 2x+1 = 3x−2 15) 6y+2 = 5y
16) 7x−3 = 3x+1 17) 62x+1 = 5x+2 18) 91−x = 12x+1
19) 52x−1 = 3x−3 20) 3x−2 = 42x+1 21) 83x−2 = 9x+2
22) 22x−3 = 5−x−1 23) 103x+2 = 5x+3 24) 53x = 3x+4
25) 3x+4 = 21−3x 26) 42x+3 = 5x−2 27) 32−3x = 42x+1
28) 22x−3 = 5x−2

3.4 Solving Logarithmic Equations
In the previous section, we took exponential equations and used the properties of
logarithms to restate them as logarithmic equations. In this section, we will take
logarithmic equations and use properties of logarithms to restate them as expo-
nential equations. In the previous section, we used the property of logarithms
that said logb M p = p logb M . In this section, we will make use of two additional
properties of logarithms:
M
logb (M ∗ N ) = logb M + logb N and logb N
= logb M − logb N
Just as our previous property of logarithms was simply a restatement of the rules
of expoenents, these two properties of logarithms depend on the rules of expo-
nents as well. Since we’re interested in logb M and logb N , let’s restate these in
terms of exponents:
If logb M = x then bx = M and if logb N = y then by = N
The properties of logarithms we’re interested in justifying have to do with M ∗ N

and M
N
, so let’s look at those expressions in terms of exponents:
M ∗ N = bx ∗ by = bx+y
and
M bx
= y = bx−y
N b
If we’re interested in logb (M ∗ N ), then we’re asking the question ”What power
do we raise b to in order to get M ∗ N ?” We can see above that raising b to the
x + y power gives us M ∗ N . Since x = logb M and y = logb N then x + y =
logb M + logb N , so:
logb (M ∗ N ) = x + y = logb M + logb N
Likewise, if we’re interested in logb MN

, we’re asking the question ”What power
do we raise b to in order to get N ?” Since raising b to the x − y power gives us M
M
N
and x − y = logb M − logb N , then:
M
logb N
= x − y = logb M − logb N
3.4. SOLVING LOGARITHMIC EQUATIONS 169
Let’s look at an example to see how we’ll use this to solve equations:
Example
Solve for x.
log2 x + log2 (x − 4) = 2
The first thing we can do here is to combine the two logarithmic statements into

one. Since logb (M ∗ N ) = logb M + logb N , then log2 x + log2 (x − 4) = log2 x(x − 4) .
log2 x + log2 (x − 4) = 2

log2 x(x − 4) = 2
Then we’ll restate the resulting logarithmic relationship as an exponential rela-

tionship:
22 = x(x − 4)
4 = x2 − 4x
0 = x2 − 4x − 4
4.828, −0.828 ≈ x
Most textbooks reject answers that result in taking the logarithm of a negative
number, such as would be the case for x ≈ −0.828. However, the logarithms of
negative numbers result in complex valued answers, rather than an undefined
quantity. For that reason, in this text, we will include all answers.
If a problem involves a difference of logarithms, we can use the other property of

logarithms introduced in this section.
Example
Solve for x.
log(5x − 1) − log(x − 2) = 2
Again, our first step is to restate the difference of logarithms using the property
logb M
N
= logb M − logb N :
log(5x − 1) − log(x − 2) = 2

5x − 1
log =2
x−2
We’re working with a logarithm in base 10 in this problem, so in our next step
we’ll say:

5x − 1
log =2
x−2
5x − 1
= 102
x−2
Then multiply on both sides by x − 2:
5x − 1
102 =
x−2
5x − 1
(x − 2) ∗ 100 = ∗
(x−2)

(x
−2)
And, solve for x
100x − 200 = 5x − 1
95x = 199
199
x=
95
In some equations, all of the terms are stated using logartihms. These equations
often come out in a form that says logb x = logb y. If this is the case, we can then
conclude that x = y.
It seems reasonable that if the exponent we raise b to in order to get x is the same
exponent that we raise b to in order to get y, then x and y are the same thing.
Assume:
logb x = logb y
Let’s say that logb x = a = logb y
then
ba = x and ba = y
if both x and y are equal to ba , then x = y
Example
Solve for x.
log5 (4 − x) = log5 (x + 8) + log5 (2x + 13)

First, let’s use the properties of logarithms to restate the equation so that there is
only one logarithm on each side.
log5 (4 − x) = log5 (x + 8) + log5 (2x + 13)

log5 (4 − x) = log5 (x + 8)(2x + 13)
Then, we’ll use the property of logarithms we just discussed:
If logb x = logb y
then
x=y

log5 (4 − x) = log5 (x + 8)(2x + 13)
4 − x = (x + 8)(2x + 13)
4 − x = 2x2 + 29x + 104
0 = 2x2 + 30x + 100
0 = 2(x + 5)(x + 10)
−5, −10 = x
Exercises 3.4
Solve for the indicated variable in each equation.
1) log3 5 + log3 x = 2 2) log4 x + log4 5 = 1
3) log2 x = 2 + log2 3 4) log5 x = 2 + log5 3
5) log3 x + log3 (x − 8) = 2 6) log6 x + log6 (x − 5) = 1
7) log(3x + 2) = log(x − 4) + 1 8) log(x − 1) − log x = −0.5
9) log2 a + log2 (a + 2) = 3 10) log3 x + log3 (x − 2) = 1
11) log2 y − log2 (y − 2) = 3 12) log2 x − log2 (x + 3) = 2
13) log3 x + log3 (x + 4) = 2 14) log4 u + log4 (u + 1) = 1
15) log 5 + log 5 = log 6 16) ln x + ln 4 = ln 2
17) log7 x − log7 12 = log7 2 18) log 2 − log x = log 8

19) log3 x − log3 (x − 2) = log3 4 20) log6 2 − log6 (x − 2) = log6 9
21) log4 x − log4 (x − 4) = log4 (x − 6)
22) log9 (2x + 7) − log9 (x − 1) = log9 (x − 7)
23) 2 log2 x = log2 (2x − 1)
24) 2 log4 y = log4 (y + 2)
25) 2 log(x − 3) − 3 log 2 = 1
26) 2 log5 7 − log5 (x + 1) = log5 (2x − 5)

3.5. APPLICATIONS OF THE NEGATIVE EXPONENTIAL FUNCTION 175
3.5 Applications of the Negative Exponential Func-

tion
At the beginning of Chapter 3, we worked with application problems and solved

them using the graphing calculator. In this section, we will revisit some of these
application problems and use the solution methods discussed in the previous
sections to solve these problems algebraically.
Radioactive Decay
The decay of a radioactive element into its non-radioactive form occurs following
a time line dictated by the ”half-life” of the element. The half-life is the amount
of time that it takes for half of the existing radioactive material to decay to its
non-radioactive form.
Consider the equation:
A(t) = A0 e−kt
where A(t) is the amount of material left at time t, A0 is the amount present at
t = 0, and k is a constant that can be determined based on the half-life of the
material.
If we know that after one half-life, there will 50% of the radioactive material re-
maining, then we can say that:
0.5A0 = A0 e−kth
where th is the half-life. To solve this equation for k, we would first divide on
both sides by A0 :
0.5A0 A0 e−kth
=
A0 A0
0.5 = e−kth
Then take the natural logarithm of both sides and bring the exponent down in
front of the expression as a coefficient:
0.5 = e−kth
ln(0.5) = ln(e−kth )
ln(0.5) = −kth ∗ ln(e)
ln(0.5) = −kth ∗ 1
ln(0.5)
− =k
th
The value of k can then be used in the equation A(t) = A0 e−kt to determine the
amount of material left after any time t.
Example
The isotope Gold-198 (198 Au) is a type of gold sometimes used in medical appli-
cations and has a half-life of 2.7 days. How much of a 65 gram sample of 198 Au
will be left after 6 days? How long would it take for there to be 10 grams left?
If we know the half-life, we can calculate the value of the constant k.
ln(0.5)
k=−
th
ln(0.5)
k=−
2.7
k ≈ 0.2567
Now that we know the value of k, we can directly calculate the amount of 198 Au
left after 6 days:
A(t) = A0 e−kt
= 65e−0.2567∗6
≈ 13.932
So, approximately 13.932 grams of 198 Au would be left after 6 days.
In order to calculate how long it takes for 10 grams of 198 Au to be left, we’ll need
to solve for t in the equation A(t) = A0 e−kt with A(t) = 10:
A(t) = A0 e−kt
10 = 65e−0.2567t
First, we’ll divide on both sides by 65:
10 −0.2567t
65e
=
65 65

10
= e−0.2567t
65
Then, take the natural logarithm on both sides:

10
= ln e−0.2567t

ln
65
We’ll calculate an approximate value for ln 10

65
and restate the right hand side of
the equation using the properties of logarithms:
−1.872 ≈ −0.2567t ∗ ln e
−1.872 ≈ −0.2567t ∗ 1
−1.872 ≈ −0.2567t
7.3 ≈ t
So, it would take about 7.3 days for the 65 grams of 198 Au to decay to 10 grams.
Newton’s Law of Cooling

Newton’s Law of Cooling states that the temperature of an object can be deter-
mined using the equation:
T = Ta + Ce−kt
where Ta is the ambient temperature of the surrounding environment. The values

of the constants C and k can often be calculated from given information.
Example
A bottle of soda at room temperature (72◦ F) is placed in a refrigerator where the

temperature is 44◦ F.
After half an hour, the soda has cooled to 61◦ F. What is the temperature of the
soda after another half hour?
First, since the soda is placed in the refrigerator where the ambient temperature is
44◦ F, then Ta = 44. We also know that at t = 0, T = 72, which is the temperature of
the soda can when it is first put in the refrigerator. This will allow us to calculate
the constant C.
T = Ta + Ce−kt
72 = 44 + Ce−k∗0
72 = 44 + Ce0 = 44 + C ∗ 1
72 = 44 + C
28 = C
Now we know that Ta = 44 and C = 28. We can use the other piece of information
from the problem to calculate the value of k. The problem states that after 30
minutes the soda has cooled off to 61◦ F. That means that when t = 30 (or t = 0.5
depending on which units you choose) the temperature T will be 61◦ F. We can
then set up the equation to reflect this and calculate the vlaue of k:
T = 44 + 28e−kt
61 = 44 + 28e−k∗30
First, we’ll subtract 44 on both sides and divide by 28:
61 = 44 + 28e−k∗30
17 = 28e−30k
17
= e−30k
28
Now, we’ll take the natural logarithm on both sides to bring the −30k down from
the exponent:

17
= ln e−30k

ln
28

17
ln = −30k ∗ ln e
28
−0.499 ≈ −30k
0.01663 ≈ k
Now, we have the full formula for calculating temperature in this scenario:
T = 44 + 28e−0.01663t
To find out what happens after 60 minutes, we can simply plug in 60 for t:
T = 44 + 28e−0.01663t
T = 44 + 28e−0.01663∗60
T = 44 + 28e−0.9978
T ≈ 44 + 28 ∗ 0.3687
T ≈ 44 + 10.3236 ≈ 54.3◦ F
Exercises 3.5
1) If the half-life of radioactive cesium-137 is 30 years, find the value of k in the

equation A(t) = A0 e−kt .
a) Given a 10 gram sample of cesium-137, how much will remain after 80

years?
b) How long will it take for only 2 grams of cesium-137 to remain?
2) The half-life for radioactive thorium-234 is about 25 days. Use this to find
the value of k in the equation A(t) = A0 e−kt .
a) How much of a 40 gram sample will remain after 60 days?
b) After how long will only 10 grams of thorium-234 remain?
3) Given a sample of strontium-90, it is known that after 18 years there are

32mg remaining and after 65 years there are 10mg remaining. Use this informa-
tion to find out how much strontium-90 was in the sample to begin with, and also
determine the half-life of strontium-90.
4) A 12 mg sample of radioactive polonium decays to 7.26 mg in 100 days.
a) What is the half-life of polonium?
b) How much of the 12 mg sample remains after 180 days?

5) A hot bowl of soup is served at a dinner party. It starts to cool according to

Newton’s Law of Cooling so that its temperature at time t is given by:
y = 65 + 145e−0.05t
where t is measured in minutes and y is measured in degrees Fahrenheit.
a) What is the initial temperature of the soup?
b) What is the temperature after 10 minutes?
c) After how long will the temperature be 100◦ ?
6) Newton’s Law of Cooling is used in homicide investigations to determine

the time of death. The normal body temperature is 98.6◦ F. Immediately following
death, the body begins to cool. This process uses Newton’s Law of Cooling:
y = Ta + Ce−kt
If the ambient temperature is 60◦ , and the body has cooled to 72◦ F after 6 hours,
use this information to determine the value of k in the equation.
7) The police discover the body of a murder victim. Critical to solving the
crime is determining when the murder was committed. The coroner arrives at
the murder scene at 12 Noon. She immediately takes the temperature of the body
and finds it to be 94.6◦ F. She then takes the temperature 1.5 hours later and finds
it to be 93.4◦ F. If the temperature of the room is 70◦ F, when was the murder com-
mitted?
8) A roasted turkey is taken from the oven when its temperature has reached
185 F and is placed on a table in a room where the temperature is 75◦ F.
◦
a) If the temperature of the turkey is 150◦ F after 30 minutes, what is its tem-
perature after 45 minutes?
b) When will the turkey cool off to 100◦ F?
9) A kettle full of water is brought to a boil in a room with an ambient temper-

ature of 20◦ C. After 15 minutes, the temperature of the water has decreased from
100◦ C to 75◦ C. Find an equation using Newton’s Law of Cooling to represent to
temperature at time t. Find the temperature of the water after 25 minutes.
10) A cup of coffee with a temperature of 105◦ F is placed in a freezer with a

temperature of 0◦ F. After 5 minutes, the temperature of the coffee is 70◦ F. Find an
equation using Newton’s Law of Cooling to represent to temperature at time t.
What will the temperature be in 10 minutes?
Answer Key
Section 1.1
1) −x2 − 16x − 13 3) 3b2 5) 6m − 3
7) −2a2 + 19a − 58 9) 2y 2 + 24y − 22
Section 1.2
1) 8a2 b2 (b + 3) 3) 13t2 (t6 + 2t2 − 3)
5) 9mn3 (5m3 n2 + 4n3 + 9m) 7) (a + 2)(a + 1)
9) (x − 9)(x + 3) 11) (m + 9)(m − 6)
13) (a + 3)(a − 3) 15) (k + 7)(k − 7)
17) 6(x + 3)(x − 3) 19) 2(10 + a)(10 − a)
21) 2(7 + k)(7 − k) 23) 5(y + 4)(y − 4)
25) 2(2y + 7)(2y − 7) 27) k(6 + 7k)(6 − 7k)
185
Section 1.2(cont.)
29) PRIME 31) (3x − 4)(3x − 2)
33) (2x + 3)(x + 2) 35) 2(5y + 3)(2y + 1)
37) 3(2a − 3)(4a − 1) 39) (10 − y)(3 + y)
41) (12 + x)(2 − x) 43) (14 + x)(6 − x)
45) 3(2y 2 + 8y + 5) 47) 4a(5x + 1)(x − 2)
Section 1.3
1) x ≈ 0.275, −7.275 3) x ≈ 1.587, −0.420
5) x ≈ 0.787, −1.905 7) x ≈ −0.548, 2.002
9) x ≈ −0.164, −1.172 11) x ≈ −0.575, −2.175
13) x = 10, 5
Section 1.4
Im
10i
9i
8i
7i #11
#3 6i
5i #1
#9 4i
3i
2i
1i
#5
R
−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10
−1i
−2i #7
−3i
−4i
−5i
−6i
−7i
−8i
−9i
−10i
13) 6i 15) 10i 17) 2i 19) 1.414i
21) 3.162i 23) 2.236i

Section 1.4(cont.)
25) 11 + 10i 27) 8 + 10i 29) 2−i 31) −42
33) 10 35) 1 + 5i 37) 27 − 28i 39) 11 + i
41) 12 − 16i 43) 10 45) 97
47) −i 49) i 51) i 53) 1
55) −1 57) 1 59) −i 61) −1
Section 1.5
1) x ≈ 1.758, −0.758 3) x ≈ 0.344, −1.744 5) x ≈ 0.16 ± 0.373i
7) x ≈ −0.5 ± 1.204i 9) x ≈ 0.286 ± 0.881i 11) x ≈ 4 ± 4.243i
Section 1.6
3 x 4x
1) 3) − 5)
x−3 3 2x − 1
y+5 x+5
7) − 9) 11) −(x + 7)
2(y + 1) x−1
3 2a + 3 1
13) 15) 17)
5x(x − 2) a+1 x−y
Section 1.6 (cont.)
4x
19) 21) 3x
(x − 2)(x − 3)
(x + 1)(x − 2) (x − 4)(x + 3)
23) 25)
(x − 1)(x − 2) (x − 1)(2x + 1)
Section 1.7
1 6x − 6 −k 2 + 5k + 23
1) 3) 5)
x(x − 1) x2 − 9 (k + 2)(k + 5)
−y 2 − 3y x2 − 2x + 1 b−1
7) 9) 11)
y 2 − 25 (x + 1)(x − 3) 2b + 2
1 −1 x
13) 15) 17)
a+1 2x − 4 x−2
a 5x + 3 5
19) 21) 23)
a−3 x+1 x−1
Section 1.8
2 m2 x
1) 3) 5)
2x + y n(m − n) y
x+2 −5
7) x+1 9) 11)
x−2 x2 −x−5
Section 1.8 (cont.)
2b − a (n + 3)(n − 2) −x
13) − 15) 17)
3b − a n−1 1−x
OR
x
x−1
Section 1.9
1) x = −1, −5 3) y ≈ 3.449, −1.449 5) y=7
7) x = 6, −4 9) n = 12, −2 11) x = −0.2, 3
13) x ≈ −2.256, −5.911 15) x = 0, 16 17) x ≈ 0.884, −1.884
19) y ≈ 5.919, −2.253 21) a = 0, 32 23) x = −0.625
3
25) y= 7
27) x ≈ 13.659, −0.659 29) x=1
Section 2.1
1) −7 < x ≤ −4 OR 0 ≤ x ≤ 5 3) x ≤ −1 OR x ≥ 2
5) −8 ≤ x < −6 OR x ≥ 3 7) x ≤ −10 OR −4 < x < 1 OR x > 6
9) −12 ≤ x < −7 OR x > 5 11) x = 0 OR x > 4
13) x < −5 OR −3 < x < −1 OR x ≥ 4

Section 2.2
1) x ≈ −0.4142, 2.4142, 31 3) x ≈ −4.035, 2.378, 2.657
5) x ≈ 2.176, 2.902, −0.902, −0.176 7) x ≈ −1.984, 0.707, 1.286, 4.991
9) x ≈ 6.161, −0.537 11) x ≈ −0.889, 1, 0.645, 1.745
Section 2.3
1) y ≥ 0: −0.618 ≤ x ≤ 1.618 OR x ≥ 3
y < 0: x < −0.618 OR 1.618 < x < 3
3) y ≥ 0: −1.5 ≤ x ≤ −0.236 OR x ≥ 4.236
y < 0: x < −1.5 OR −0.236 < x < 4.236
5
5) y ≥ 0: x ≤ 1.5 OR x ≥ 3
5
y < 0: 1.5 < x < 3
7) y ≥ 0: x ≤ −2.656 OR −0.486 ≤ x ≤ 0.460 OR x ≥ 1.682
y < 0: −2.656 < x < −0.486 OR 0.460 < x < 1.682
9) x ≤ −3 OR x = 1 11) x > 3 OR −4 < x < 1
13) All real numbers 15) x ≤ −2.264 OR 0.756 ≤ x ≤ 3.508
17) x < 0.5 OR x > 0.839

Section 2.4
1) −4 < x < 2 OR x > 6
3) −5 < x < −2 OR 2 < x < 7
5) − 32 < x < 1.791 OR x < −2.791
7) −2.080 < x < −1.618 OR x > 0.618
9) − 11
3
< x < −2.162 OR x > 4.162
11) −2 < x < 1.791 OR x < −2.791 OR x > 3
13) −4.372 < x < −3.828 OR 1.372 < x < 1.828
15) −5.193 < x < −2.646 OR 0.193 < x < 2.646
Section 2.5
1) x2 + 3x − 4 = 0 3) 2x2 − 5x + 3 = 0
5) 3x2 − 10x + 3 = 0 7) 4x2 − 12x − 7 = 0
9) 3x2 + 11x + 6 = 0 11) 2x2 − x + 15 = 0
Section 2.6
1) y 2 − 2y + 2 3) x−5 5) x3 − x + 3
7) (2z 2 − 2z + 7) R:1 9) 4x2 + x + 1
11) (2y 3 − 3y 2 − 2y + 2) R : 3y + 2
13) (5x2 + 15x + 17) R : −24x − 83

Section 2.7
1) x2 − 3x − 10 3) x2 + 5x − 1 5) x3 − 4x2 + x + 6
7) x3 − 1 R:8 9) x3 + 1 11) x3 − 3x2 + 4x − 2
13) x2 + 4x + 5
Section 2.8
1) (x + 1)(x − 2)(x2 − 2x + 5) 3) (x + 1)(x − 2)(2x2 − 3x + 2)
x = −1, 2, 1 ± 2i x = −1, 2, 0.75 ± 0.661i
5) (x − 5)(x + 2)(x2 − 6x + 13) 7) (x − 1)3 (x2 − 6x + 10)
x = 5, −2, 3 ± 2i x = 1, 3 ± i
9) (x − 1)3 (x − 3) 11) (x − 1)3 (x2 + 9)
x = 1, 3 x = 1, ±3i
13) (5x + 1)(3x2 − 2x + 3) 15) (3x + 2)(2x2 + 3x + 2)
x = − 51 , 13 ± 0.943i x = − 32 , −0.75 ± 0.661i
17) (x + 3)(2x − 1)(2x2 + 5x + 5) 19) (2x + 1)(x − 3)(x2 + x + 1)
x = −3, 21 , −1.25 ± 0.968i x = 3, − 21 , −0.5 ± 0.866i
21) (2x + 1)(3x − 2)(2x2 + x + 1) x = 23 , − 12 , −0.25 ± 0.661i

Section 3.1
1) a) ≈ 107mg 3) a) 450 bass

b) ≈ 99 minutes b) ≈ 744 bass
5) a) ≈ 141◦ F 7) a) 13 kg.
b) ≈ 28.3 minutes b) ≈ 6.62 kg.
c) ≈ 63.7 days
9) a) 6,000 10,570 11) a) 33,333 21,494

b) 1.6 years 12,000 b) ≈ 1.98 years
approaches 20,000
13) a) 16.5 lbs 15) a) 6.8 gal.

b) 45.8 minutes 43.94 minutes
Section 3.2
1) 5t = 9 3) 52 = 25 5) 10−1 = 0.1
7) 100.845 ≈ 7 9) 25.13 ≈ 35 11) e−1.3863 ≈ 0.25
1 3
13) log 100 = 2 15) log4 1024
= −5 17) log16 8 = 4
19) log 20 ≈ 1.3 21) ln 20.0855 ≈ 3 23) ln 0.0183 ≈ −4

Section 3.3
1) ≈ 2.322 3) ≈ 1.771 5) ≈ 1.585 7) ≈ 1.227
9) ≈ −0.089 11) ≈ 0.183 13) ≈ −2.297 15) ≈ −19.655
17) ≈ 0.723 19) ≈ −0.795 21) ≈ 2.117 23) ≈ 0.042
25) ≈ −1.165 27) ≈ 0.134
Section 3.4
1) x = 1.8 3) x = 12 5) x = 9, −1
16
7) x=6 9) x = 2, −4 11) x= 7
13) x ≈ 1.606, −5.606 15) x = 1.2 17) x = 24
8
19) x= 3
21) x = 3, 8 23) x=1
25) x ≈ 11.944, −5.944
Section 3.5
1) a) ≈ 1.576g b) ≈ 69.67 years 3) 50mg
5) a) 210◦ F b) ≈ 152.9◦ F c) ≈ 28.4 minutes
7) 7:30AM 9) a) T = 20 + 80e−0.025t b) ≈ 62.85◦ C

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1.1 Algebraic Simplification . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.4 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1.5 Quadratic Equations with Complex Roots . . . . . . . . . . . . . . . 43

1.6 Multiplying and Dividing Rational Expressions . . . . . . . . . . . 50

1.7 Adding and Subtracting Rational Expressions . . . . . . . . . . . . 58

1.8 Complex Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

1.9 Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

2 Polynomial and Rational Functions 83

2.1 Representing Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . 83

2.2 Solution by Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

2.3 Solution of Polynomial Inequalities by Graphing . . . . . . . . . . . 92

2.4 Solution of Rational Inequalities by Graphing . . . . . . . . . . . . . 100

2.5 Finding Factors from Roots . . . . . . . . . . . . . . . . . . . . . . . 108

2.6 Polynomial Long Division . . . . . . . . . . . . . . . . . . . . . . . . 111

2.7 Synthetic Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

2.8 Roots and Factorization of Polynomials . . . . . . . . . . . . . . . . 128

3 Exponents and Logarithms 135

3.1 Exponential and Logistic Applications . . . . . . . . . . . . . . . . . 135

3.2 Logarithmic Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

3.3 Solving Exponential Equations . . . . . . . . . . . . . . . . . . . . . 161

3.4 Solving Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . 168

3.5 Applications of the Negative Exponential Function . . . . . . . . . 175

Answer Key 185

1.1 Algebraic Simplification

3(x − 1)(2x + 5) − (x + 4)2

3(x − 1)(2x + 5) − (x + 4)2 = 3(2x2 + 3x − 5) − (x2 + 8x + 16)

In the case of (2x2 + 3x − 5), there is a 3 which must be distributed, resulting in

3(x − 1)(2x + 5) − (x + 4)2 = 3(2x2 + 3x − 5) − (x2 + 8x + 16)

2(x + 3)2 − 4(3x − 1)(x + 2)

2(x + 3)2 − 4(3x − 1)(x + 2) = 2(x2 + 6x + 9) − 4(3x2 + 5x − 2)

Or, we can choose to distribute the 4 first:

2(x + 3)2 − 4(3x − 1)(x + 2) = 2(x2 + 6x + 9) − (12x − 4)(x + 2)

2(x + 3)2 − 4(3x − 1)(x + 2) = 2(x2 + 6x + 9) + (−12x + 4)(x + 2)

2(x + 3)2 − 4(3x − 1)(x + 2) = 2(x + 3)(x + 3) − 4(3x − 1)(x + 2)

3x[5 − (2x + 7)] + (3x − 2)2 − (x − 5)(x + 4)

3x[5 − (2x + 7)] + (3x − 2)2 − (x − 5)(x + 4)

= 3x[5 − 2x − 7] + (3x − 2)2 − (x − 5)(x + 4)

Simplify each expression.

1) (x − 2)[2x − 2(3 + x)] − (x + 5)2

2) 3x2 − [7x − 2(2x − 1)(3 − x)]

3) (a + b)2 − (a + b)(a − b) − [a(2b − 2) − (b2 − 2a)]

4) 5x − 3(x − 2)(x + 7) + 3(x − 2)2

6) (a − 1)(a − 2) − (a − 2)(a − 3) + (a − 3)(a − 4)

7) 2a2 − 3(a + 1)(a − 2) − [7 − (a − 1)]2

8) 2(x − 5)(3x + 1) − (2x − 1)2

9) 6y + (3y + 1)(y + 2) − (y − 3)(y − 8)

10) 6x − 4(x + 10)(x − 1) + (x + 1)2

Greatest Common Factor

Factor 7x2 + 14x

7x2 + 14x = 7x(x + 2)

Factor 42x2 y 6 + 98xy 3 − 210x3 y 2

42x2 y 6 + 98xy 3 − 210x3 y 2 = 2xy 2 ∗ 21xy 4 + 2xy 2 ∗ 49y − 2xy 2 ∗ 105x2

2xy 2 (21xy 4 + 49y − 105x2 ) = 2xy 2 (7 ∗ 3xy 4 + 7 ∗ 7y − 7 ∗ 15x2 )