M 101 Unit 1
M 101 Unit 1
M 101 Unit 1
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LEARNING OBJECTIVES:
REFERENCES:
This lesson begins with the review of algebraic expression. This includes terms used in algebra
such as constants and variables, exponents, terms, and so on and so forth. Basic operations in
polynomials and synthetic division will be discussed in this lesson.
Activity 1
a. 2 b. x c. √ d. e. -4
When an algebraic expression is written as the sum of other algebraic expression, each of the
expression is called a term. Example (e) above can be written as the sum
Any sum of monomials is called a polynomial. A polynomial consisting of two terms is called a
binomial, and a polynomial of three terms is called a trinomial. Some examples of binomials are 2x – 1,
2x2 + y, 6 – y3. Note that the exponents of the variables are all positive integers. Some examples of
trinomials are 2x2 – 5x + 3, a + 2b + 3c, and 4 – xy + x2y3. Example (e) cannot be considered a trinomial
because of the term which is equal to is not a monomial. Can you tell why? A polynomial 3x – 4y
+ 7xy – 8 is a polynomial of four terms.
We may name polynomials using capital letters. A polynomial x can be represented by P(x).
Thus, we may say P(x) = 5x2 + 2x – 1. In general, we evaluate polynomials or algebraic expressions by
substituting a given value or values for all the variables found in the expression. If x = -3, then P (-3) is
the value of the polynomial evaluated at x = -3.
The degree of a term in one variable is the exponent of the variable of that term. For
example, 2x3 has a degree of 3. Any nonzero constant term has a degree of zero. The constant 0 has no
degree. The degree of a term of two or more variables is the sum of the exponents of all the variables
in the term. For instance, the term 8ab has a degree of 2. Note that both a and b has an exponent of 1
and hence, the sum of their exponents is 2. The term -4x3y2z has a degree of 6. What do you think is the
degree of a monomial 6xyz?
The degree of a polynomial is equal to the degree of the term with the highest degree. For
example, the polynomial 2x3 – 5x2 + 4 has a degree of 3 while the polynomial 8x 4y2 +
2 3 4 2 2 3
5xy + x y has a degree of 6. As a polynomial in variable x, 8x y + 5xy + x y has a degree of 4 while as
a polynomial in y, it has a degree of 2.
P(x) =
a. 6x2y e. -5
b. -3y4 f. 3x4 – y3
c. 2y – 1 g. 7x2 – 3x + 10x5
d. 4x2y3 – 2xy2 + x3 h. 0
Solution:
a. The expression 6x2y is a monomial and its degree is 3.
b. The expression -3y4 is a monomial and its degree is 4.
c. The expression 2y – 1 is a binomial and its degree is 1. Note that y1 = y.
d. The expression 4x2y3 – 2xy2 + x3 is a trinomial and its degree is 5. Note that among
the terms, 4x2y3 has the highest degree. The variable x has an exponent of 2 while
y has an exponent of 3, thus 2 + 3 = 5.
e. The expression -5 is a monomial. It is a constant and therefore has a degree of 0.
Note that we can multiply -5 by any variable raised to the zeroth power since any
quantity raised to the zeroth power is equal to 1. Thus, -5 can be written as -5x0
where x 0, that is -5 = -5(1) = -5x0.
f. The expression 3x4 – y3 is a binomial. As a polynomial in x, it has a degree of 4 and
as a polynomial in y, it has a degree of 3.
e. The expression 7x2 – 3x + 10x5 is a trinomial and its degree is 5.
g. The expression 0 is a monomial. It has no degree.
Activity 2
Addition of Polynomials
To find the sum of two or more polynomials, we simply combine like terms, add their numerical
coefficients and copy the common literal coefficients. Here we observe the rule on adding signed
numbers.
Activity 3
Subtraction of Polynomials
To subtract one polynomial from another polynomial, we simply change the sign of the
subtrahend and add the polynomials following the rule in adding signed numbers.
Activity 4
Multiplication of Polynomials
When a number is taken as a factor and is multiplied by itself several times, the product is
called the power of the repeated factor.
The quantity of is called the power of is the base and n is the exponent of the power.
The laws of exponents and the distributive axioms allow us to perform multiplication of
algebraic expression.
Examples:
a.
b.
=
=
c.
=
=
Activity 5
Division of Polynomials
1. = =
2. = =
= {
Examples:
a. =
b. = =
c. =
In example (c), when the base is raised to the zero power, it is equal to 1. This is true for any
base except zero.
To divide polynomials, divide the coefficient of the dividend by the coefficient of the divisor,
then apply the law of exponents for the variables to get the final quotient.
Examples:
a. Divide
b. Find the quotient when
c. Divide
Solution:
a. = =
___________ _
b. /
(-)
(-)
- 2x – 3
(-) - 2x – 3
0
Observe that the procedure is very similar to the ordinary division algorithm. In the example
above, we say that is exactly divisible by 2x + 3 because the remainder is 0.
_ ___________________
c. /
(-)
-
(-)
(-)
- 185x + 95
To check if the quotient is correct, we multiply it by the divisor. When there is a remainder, we
add it to the result and compare the sum with the dividend. This is left for you to do.
Remember that numbers multiplied to form a product are called factors of such product.
Unless otherwise indicated, we will deal only with integral facctors, that is, integers factored over the set
of integers.
This module discusses the types of special products. It also touches on factoring binomials,
trinomials, and polynomials of more than three numbers.
Some exercises are presented to help you understand the topics. Worksheets and chapter test
are provided for you to answer and will be returned on the agreed date. These exercises pose as a
challenge to develop diligence, patience, accuracy, and to enhance your analytical abilities.
There are certain products of polynomials called special products that occur frequenctly and
should be recognized. In these special products, x and y represent variables and a, b, and c represent
constants. The products can be verified by performing the multiplications.
Special Products
1. The first term is the prodcut of the first terms of the binomials.
2. The middle term is obtained by adding the product of the outer terms to the
product of the inner terms.
3. The last term is the product of the last terms of the binomials.
Examples:
(x + y)(x – y) = x2 – y2 The product of the sum and difference of two terms is the
square of the firts term minus the square of the second term.
The product of the sum and difference of the same two terms is a special case in the sense that
its middle term is zero and leads to a very simple product called the difference of two squares.
Examples:
(x + y) (x + y) = x2 + 2xy + y2 (x – y) (x – y) = x2 – 2xy + y2
The square of the sum (or the difference) of two terms is the square of the first, plus (or minus)
twice the product of the two terms plus the square of the second term.
Examples:
The cube of the sum (or the difference) of two terms is the cube of the first term plus (or minus)
thrice the square of the first times the second plus thrice the product of the first and the square of the
second plus (or minus) the cube of the second term.
There are always four terms in the cube of a binomial. All terms in the product are positive in
the cube of the sum of two terms. The cube of the difference of a binomial will have an alternate –
and + signs.
Examples:
When the trinomial factor (based on the binomial factor) is the square of the first plus (or
minus) the product of the first and the second plus the square of the second, the product is the sum (or
difference) of two cubes.
This case is so special in the sense that the definite forms of the binomial and trinomial lead to a
very simple product called the sum or difference of two cubes.
Examples:
Factoring
In every term of a polynomial contains a common monomial factor, then, by the distributive
law, the polynomial can be written as the product of the common monomial factor and the quotient
obtained by dividing the original polynomial by the common factor. For instance, a is a common
monomial factor of each term of the trinomial ax + ay + az; thus
ax + ay + az = a (x + y + z)
Examples:
a) To factor the polynomial 6x3y2 – 3x2y + 9xy, we first observe that 3xy is a common
monomial factor of each term; hence
6x3y2 – 3x2y + 9xy = 3xy (2x2y – x + 3)
x2 - y2 = (x + y) (x – y) The left side of this formula is the difference of two squares, and
the formula states that it can be written as the product of the sum and difference of square roots of
the two squares.
The difference of two squares is the prodcut of two binomials, one binomial expressing the sum
and the other binomial, the difference of the sum of two terms.
Examples:
b) x2 – 25 = (x + 5)(x – 5)
c) 9a4 – 36b2 = (3a2)2 – (6b)2 = (3a2 + 6b)(3a2 – 6b)
Activity 3 defines the square of a binomial as a trinomial which has two of its as perfect squares
and the other term as twice the product of the square roots of the two squares. This special trinomial is
called a perfect trinomial square. Given a perfect trinomial square, therefore, its factors are two
identical binomials which are either the sum (or difference) of the square roots of the two squares
depending on whether the sign of the term which is not a square is positive (or negative). In symbolic
notations.
x2 + 2xy + y2 = (x + y)2 = (x + y) (x + y)
x2 - 2xy + y2 = (x - y)2 = (x - y) (x - y)
Examples: Factor
a) x2 – 4x + 4
b) x2 + 14xy + 49y2
c) 4x2 + 12x + 9
d) 9a2 – 30ab + 25b2
Solutions:
a) In the given trinomial are two squares, x2 and 4, whose square roots are x
and 2, respectively. Since twice the product of x and 2 is the middle term of the
trinomial, x2 – 4x + 4 is a perfect square. Therefore, x2 – 4x + 4 = (x – 2) (x – 2) or
(x -2)2.
This section deals with techniques and patterns for factoring trinomials of the form x2 + bx + c,
whose quadratic coefficient is one and whose constant term is either positive or negative.
Examples:
a) (x + 3) (x + 4) = x2 + 7x + 12.
The coefficient of x is 7 which is the sum of 3 and 4.
The constant term is 12 which is the product of 3 and 4.
b) (x – 5) (x – 3) = x2 – 8x + 15
The coefficient of x which is -8 which is the sum of -5 and -3.
The constant term is 15 which is the product of -5 and -3.
The preceding examples suggest this pattern:
(x + a) (x + b) = x2 + (a + b) x + ab. By the symmetric property of equality,
it follows that x2 + (a + b) x + ab = (x + a) (x + b)
c) Factor x2 + 9x + 20
From the pattern, factors can be found by looking for two integers whose product
ab is 20 and whose sum (a + b) is 9
d) Factor x2 – 10x + 16
This time the product is 16 and the sum is – 10.
Both factors must be negative.
Negative factors of 16: -1(-16) -2(-8) -4(-4)
Corresponding sum: -17 -10 -8
2
Therefore, x – 10x + 16 = (x – 2) (x – 8)
e) (x + 8) (x – 3) = x2 + 5x – 24
The coefficient of x is 5 which is the sum of 8 and -3.
The constant term is -24 which is the product of 8 and -3.
f) (x – 7) (x + 3) = x2 – 4x – 21
The coefficient of x is -4 which is the sum of -7 and 3.
The constant term is -21 which is the product of -7 and 3.
g) Factor x2 + 8x – 20
The combination of integers must have a product of -20 and a sum of 8.
Since the sum is positive, the positive factor of -20 must have the greater absolute
value.
h) Factor x2 – 7x – 18
This time, the combination of integers must have a product of -18 and a sum of -7.
The constant number -18 suggest that the two integers must have opposite signs,
and the negative factor the greater absolute value.
Factors of -18: 1(-18) 2(-9) 3(-6)
Corresponding sum: -17 -7 -3
2
Therefore, x – 7x – 18 = (x – 9)(x + 2)
A trinomial of the form ax2 + bx + c, where the quadratic coefficient is greater that one can be
factored s binomial of the form (ax + b)(cx + d). recall the products of binomials.
Examples:
c) Factor 2x2 + 5x + 3
1) Factor: first term → 2x and x
last term → 1 and 3
Only positive factors are considered because both the middle and last terms in
the trinomial are positive.
6x 2x
3) Test the inner and outer products to find the sum 5x for the middle term.
x + 6x = 7x
3x + 2x = 5x Correct
d) Factor 2x2 – 7x + 6
1) Factor: first term → 2x and x
last term → -2 and -3; and -1 and -6
Only negative factors are considered because the middle term of the trinomial
is negative and the last term is positive.
-6x -12x
-3 x -6x
(2x - 3)(x - 2) (2x - 6)(x - 1)
-4x -2x
3) Test the inner and outer products to find the sum -7x for the middle term.
-2x + (-6x) = -8x -x + (-12x) = -13x
-3x + (- 4x) = -7x Correct -6x + (-2x) = - 8x
e) Factor 6x2 + 4x – 10
1) Factor: first term → 2x and 3x; x and 6x
last term → 1 and -10; -1 and 10; 2 and -5; -2 and 5
The last terms of the binomials must have opposite signs because the last term
in the trinomial is negative.
The first factor is the sum or difference of the cube roots of the terms in the binomial. The
second factor consists of all positive terms or alternate positive and negative terms depending on
whether a sum of two cubes or a difference of two cubes is given.
Examples:
This type is usually applied to polynomials having four or more terms. Proper grouping of terms
is necessary to make the given polunomial factorable.
Examples:
ALGEBRAIC FRACTIONS
Since division by zero is undefined, we must never permit the variable to assume values that
would cause division by zero. Any value that would cause division by zero is called an excluded value.
a)
b)
c)
d)
Solution:
𝒙 𝟐𝒚
Do not divide out the Xs in the fraction 𝒙 𝟕𝒚
. The x in the numerator is a factor of the
first term only. It is not a factor of the entire numerator. Likewise, the x in the
denominator is a factor of the first term only. It is the factor of the entire denominator.
Since the variables in a rational expression represent real numbers, the rules and procedures for
performing multiplication and division with rational expressions are the same as those for performing
these operations with arithmetic fractions. (Multiplication and division will be covered before addition
and subtraction so that you can develop some of the skills needed to add rational exponents.)
The rule in multiplying rational expressions is: If a, b, c, and d are real expressions, and b 0
and d 0, then
= The result should always be reduced to lowest terms.
1. Factor the numerator and the denominator of both rational expressions. Then write
the product as a single fraction, indicating the product of the numerators and the
product of the denominators.
2. Reduce this fraction by dividing the numerator and the denominator by any
common nonzero factors.
a) b) c) •(
Solution :
Division of rational expressions follows the same rule as in division of arithmetic fractions. That
is, if a, b, c, and d are real expressions, and b 0, c 0, and d 0, then
1.
Rewrite the division problem as the product of the dividend and the reciprocal of
the divisor.
2. Perform
Examples: Divide thethe multiplication
following using
algebraic the rules for multiplying rational expressions.
fractions.
a) b)
Solution:
The rules for adding and subtracting rational expressions are the same as those for adding and
subtracting arithmetic fractions. If a, b, and c are real numbers, and c 0, then
b)
Solution:
It is easy to make errors in the order of operations if you are not careful when adding or
subtracting rational expressions. Examples below illustrate how parentheses can be used to avoid errors.
b)
Solution:
A Word of Caution!
The – sign between the fractions in these examples affects every term of the numerator.
Whenever we subtract one fraction from another, we must remeber to subtract each term of the
numerator in the second fraction.
Remember that = = . These alterntive forms are often useful when you are working
with these two fractions whose denominators are opposites.
Example: Simplify -
Solution: - = + = = = =
Noting that the denomiator 3x – 6 and 6 – 3x are opposites, change to the equivalent form .
Now that the denominators are the same, you can add the numerators. Then factor and reduce the result by
dividing both the numerator and the denominator by 3.
Rational expressions can be added or subtracted only if they are expressed in terms of a
common denominator. Although any common denominator can be used, we can simplify our work
considerably by using the least common denominator (LCD). Sometimes the LCD can be determined
by inspection.
Activity 23
a) b) c)
d) e) 3x – 5 -
Solution:
Some of the steps illustrated in example can be combined in order to shorten the solution. Although
most of the steps in the solution are still in examples d and e, some have been combined. When
combining steps, be careful not to make an error in sign
= •
Now, convert each fraction so that it has
the LCD (m + 3)(m – 2) as its
= denominator. Simplify the numerator by
multiplying and combining like terms.
=
Factor the numerator and reduce by
dividing out the common factor m – 2.
=
Activity 24
There are two methods of simplifying complex fractions. Both are worth learning since some
problems can be worked more easily by the first method while others by the second method. Example
a illustrates method 1 and example b illustrates method 2.
Activity 25
Fractional Equations are solved by multiplying both sides of the equations by the LCD. This
method produces an equation equivalent to the original one as long as we do not multiply by zero.
Multiplying by an expression that is equal to zero can produce an equation that is not equivalent to
the original equation and thus can produce an extraneous value. An extraneous value is not a solution
of the original equation. Hence, when solving an equation with a variable in the denominator, check
that the solutions do not include a value excluded from the domain of the variable because of division
by zero.
Examples: Solve
1) 2)
Solution:
a)
Note that x – 7 is an excluded value.
( ) ( ) Multiply both members of the equations by
the nonzero LCD x – 7. The LCD is nonzero
3 + 5x – 35 = 8
since 7 is an excluded value. Solve the
5x = 40 resulting equation.
x=8
Since x = 8 is not an excluded value, this
Answer: The solution set is {8}. solution should satisfy the equation.
Verify.
b) Note that the only excluded value is a = 3.
Noting that the denominators are opposites,
( ) = (a – 3)(4) + ( ) change to .