ch12 PDF

12
Polynomials 1
TERMINOLOGY
Coefcient: A constant multiplied by a pronumeral in an
algebraic term e.g. in ax3 the a is the coefficient
Monic polynomial: A polynomial where the leading

coefficient is 1
Degree: The value of the highest power of x in a

polynomial
Polynomial: A sum or difference of terms involving

integral powers of a variable, usually x. A function of the
form P (x) = a0 + a1 x + a2 x 2 + f + an x n where a0, a1, ...
are real numbers and n is a positive integer or zero
Dividend: The number, algebraic expression or polynomial

that is being divided by another of the same type
Divisor: A number, algebraic expression or polynomial
that divides another of the same type
Factor theorem: If P(x) is divided by x - a and P (a) = 0
then x - a is a factor of P(x)
Leading term: The term with the highest power of x.
e.g. 5x 3 - 2x 2 + 3 has a leading term of 5x 3
Long division: A division of one polynomial into another
polynomial using a method similar to long division of
numbers
Quotient: The result when two numbers, algebraic

expressions or polynomials are divided
Remainder theorem: If P(x) is divided by x - a then the
remainder is given by P(a)
Root of a polynomial equation: The solution of polynomial
equation P (x) = 0. Graphically it is where the polynomial
crosses the x-axis.
Zeros: The zeros of a polynomial are the roots of the
polynomial equation P (x) = 0. They are the values that
make P(x) zero.
Chapter 12 Polynomials 1
663
INTRODUCTION
POLYNOMIALS ARE AN IMPORTANT part of algebra and are used in many
branches of mathematics. Some examples of polynomials that you have
already studied are linear and quadratic functions.
In this chapter you will study some properties of polynomials in general,
and relate polynomial expressions to equations and graphs.
DID YOU KNOW?

The word polynomial means an expression with many terms. (A binomial has 2 terms and a
trinomial has 3 terms). Poly means many, and is used in many words, for example, polyanthus,
polygamy, polyglot, polygon, polyhedron, polymer, polyphonic, polypod and polytechnic. Do you
know what all these words mean? Do you know any others with poly-?
Definition of a Polynomial
A polynomial is a function defined for all real x involving positive powers of x
in the form:
P ] x g = p 0 + p 1 x + p 2 x 2 + f + p n - 1 x n - 1 + p n x n where n is a
positive integer or zero.
P(x) is a continuous and differentiable function.

Although the definition has the term pnxn last, we generally write
polynomials from the highest order down to the lowest. e.g. f ] x g = x 2 - 5x + 4.
We can describe various aspects of polynomial as follows:
p n x n + p n - 1 x n - 1 + p n - 2 x n - 2 + f + p 2 x 2 + p 1 x + p 0 is called a polynomial
expression
P ] x g = p n x n + p n - 1 x n - 1 + p n - 2 x n - 2 + f + p 2 x 2 + p 1 x + p 0 has degree n
where p n ! 0
p n, p n - 1, p n - 2, f p 0 are called coefficients
pnxn is called the leading term and pn is the leading coefficient
p0 is called the constant term
If p n = 1, P ] x g is called a monic polynomial
If p 0 = p 1 = p 2 = f = p n = 0 then P(x) is the zero polynomial
The degree of a polynomial

is the highest power of x
with non-zero coefficient.
Coefficients can be any real

number but we generally
use integers in this course
for convenience.
664
Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES
1. Which of the following are polynomial expressions?
(a) 4 - x + 3x 2
(b) 3x 4 - x 2 + 5x - 1
(c) x 2 - 3x + x -1
Solution
(a) and (b) are polynomials but (c) is not, since it has a term x -1 that is
not a positive power of x.
2. For the polynomial P ] x g = x 6 - 2x 4 + 3x 3 + x 2 - 7x - 3
(a) Find the degree.
(b) Is the polynomial monic?
(c) State the leading term.
(d) What is the constant term?
(e) Find the coefficient of x4.
Solution
(a)
(b)
(c)
(d)
(e)
Degree is 6 since x6 is the highest power.

Yes, the polynomial is monic since the coefficient of x6 is 1.
The leading term is x6.
The constant term is -3.
The coefficient of x4 is -2.
Polynomial equation
P ] x g = 0 is a polynomial equation of degree n
The real values of x that satisfy the equation are called the real
roots of the equation or the real zeros of the polynomial.
EXAMPLES
1. Find the zeros of the polynomial P ] x g = x 2 - 5x.
Solution
To find the zeros of the polynomial, we solve P ] x g = 0.
x 2 - 5x = 0
x ]x - 5 g = 0
x = 0, x - 5 = 0
x=5
So the zeros are 0, 5.
2. Find the roots of the polynomial equation x 3 - 2x 2 - 3x = 0.
Solution
x 3 - 2x 2 - 3x = 0
x ^ x 2 - 2x - 3 h = 0
x ]x - 3 g]x + 1 g = 0
x = 0, x - 3 = 0, x + 1 = 0
x = 3,
x = -1
The roots are x = 0, 3, -1.
3. Show that the polynomial p ] x g = x 2 - x + 4 has no real zeros.
Solution
We look at the polynomial equation p ] x g = 0.
x2 - x + 4 = 0
The discriminant will show whether the polynomial has real zeros.
b 2 - 4ac = ] -1 g2 - 4 ] 1 g ] 4 g
= 1 - 16
= -15
10
So the polynomial has no real zeros.
4. For the polynomial P ] x g = ax 5 - 3x 4 + x 3 - 7x + 1
(a) Evaluate a if the polynomial is monic.
(b) Find the degree of the derivative Pl(x) .
Solution
(a) For a monic polynomial, a = 1
(b) P l^ x h = 5ax 4 - 12x 3 + 3x 2 - 7
Pl(x) has degree 4 (highest power).
665
666
12.1
1.
2.
3.
4.
Exercises
Write down the degree of each

polynomial.
(a) 5x 7 - 3x 5 + 2x 3 - 3x + 1
(b) 3 + x + x 2 - x 3 + 2x 4
(c) 3x + 5
(d) x 11 - 5x 8 + 4
(e) 2 - x - 5x 2 + 3x 3
(f) 3
(g) 2x 4 - x
For the polynomial
P(x) = x 3 - 7x 2 + x - 1, find
(a) P(2)
(b) P(-1)
(c) P(0)
Given P (x) = x + 5 and
Q (x) = 2x - 1, find
(a) P (-11)
(b) Q (3)
(c) P (2) + Q (-2)
(d) the degree of P (x) + Q (x)
(e) the degree of P (x) $ Q (x)
For the polynomial
P (x) = x 5 - 3x 4 - 5x + 4, find
(a) the degree of P (x)
(b) the constant term
(c) the coefficient of x 4
(d) the coefficient of x 2
5.
Find the zeros of the following

polynomials.
(a) P (x) = x 2 - 9
(b) p (x) = x + 5
(c) f (x) = x 2 + x - 2
(d) P (x) = x2 - 8x + 16
(e) g (x) = x 3 - 2x 2 + 5x
6.
Find the derivative of each

polynomial P (x) and state the
degree of Pl(x)
(a) P (x) = 3x 4 - 2x 3 - x 2 + 4x - 5
(b) P (x) = 5x 2 + 3
(c) P (x) = 9x 12 - 7x 5 + 8x
(d) P (x) = x 7 - 3x 3 + x 2 - 7x - 3
(e) P (x) = 8x + 5
7.
Which of the following are not

polynomials?
1
(a) 5x 4 - 3x 2 + x + x
x2 + 3x
x 2 + 3x - 7
3x + 5
0
1
3x 2 - x + 1
2
3
(g) 4x + 7x -2 + 5
(b)
(c)
(d)
(e)
(f)
8.
For the polynomial

P (x) = (a + 1) x 3 + (b - 7) x 2 + c + 5,
find values for a, b or c if
(a) P (x) is monic
(b) the coefficient of x 2 is 3
(c) the constant term is -1
(d) P (x) has degree 2
(e) the leading term has a
coefficient of 5
9.
Given P (x) = 2x + 5,
Q (x) = x 2 - x - 2 and
R (x) = x 3 + 9x, find
(a) any zeros of P (x)
(b) the roots of Q (x) = 0
(c) the degree of P (x) + R (x)
(d) the degree of P (x) $ Q (x)
(e) the leading term of Q (x) $ R (x)
10. Given f (x) = 3x 2 - 2x + 1 and

g (x) = 3x - 3,
(a) show f (x) has no zeros
(b) find the leading term of
f (x) $ g (x)
(c) find the constant term of
f (x) + g (x)
(d) find the coefficient of x in
f (x) $ g (x)
(e) find the roots of
f (x) + g (x) = 0
11. State how many real roots there
are for each polynomial equation
P ] x g = 0.
(a) P ] x g = x 2 - 9
(b) P ] x g = x 2 + 4
(c) P ] x g = x 2 - 3x - 7
(d)
(e)
(f)
(g)
P ] x g = 2x 2 + x + 3
P ] x g = 3x 2 - 5x - 2
P ]x g = x ]x - 1 g]x + 4 g]x + 6 g
P ]x g = ]x + 1 g]x - 2 g]x - 5 g
12. For the polynomial

P ] x g = 2x 3 + 3x 2 - 36x + 17,
find the roots of the derivative
polynomial equation Pl(x) = 0.
13. If P ] x g = 3x 4 - 4x 3 - 1, find the

zeros of Pl(x) .
14. Show that Pl(x) = 0 has no real
roots if P ] x g = x 3 - x 2 + 9x.
15. Show that Ql(x) = 0
has equal roots given
Q ] x g = x 3 - 3x 2 + 3x + 5.
Division of Polynomials
You would have learned how to do long division in primary school, but have
probably forgotten how to do it! We use this method to divide polynomials.
Class Investigation
Here are two examples of long division.
1. Divide 5715 by 48.
119 r3
48 g 5715
48
91
48
435
432
3
This means
3
5715
= 119 +
48
48
3
5715
# 48 = 119 # 48 +
# 48
48
48
So
5715 = 48 # 119 + 3
(check this on your calculator)
The number 5715 is called the dividend, the 48 is the divisor, 119 is the
quotient and 3 is the remainder.
2. Divide 4871 by 35.
139 r6
35 g 4871
35
137
105
321
315
6
CONTINUED
667
668
This means
4871
6
= 139 +
35
35
or
4871 = 35 #139 + 6
(check this on your calculator)
The number 4871 is called the dividend, the 35 is the divisor, 139 is the
quotient and 6 is the remainder.
Use long division to divide other numbers and write them in the form
above.
For example:
1.
2.
3.
4.
5.
2048 '
5876 '
3546 '
2992 '
8914 '
15
17
21
33
19
A polynomial P(x) can be written as P ] x g = A ] x g $ Q ] x g + R ] x g

where P(x) is the dividend, A(x) is the divisor, Q(x) is the quotient
and R(x) is the remainder.
Proof
If we divide a polynomial P(x) by A(x), we can write P(x) in the form of
P (x)
R (x)
= Q (x) +
where Q(x) is the quotient and R(x) is the remainder.
A (x)
A (x)
P (x)
R (x )
# A (x ) = Q ( x ) # A ( x ) +
# A ( x)
A (x )
A (x )
P ]x g = A ]x g $ Q ]x g + R ]x g
The division continues until the remainder can no longer be broken down
further by division.
The degree of remainder R(x) is always less than the degree
of the divisor A(x).
Proof
Suppose the degree of R(x) is higher than the degree of A(x).
This means that R(x) can be divided by A(x).
R 1(x)
R (x )
= Q 1 (x) +
A (x )
A (x )
R ] x g = A ] x g $ Q1] x g + R1] x g
So
= R2] x g
This gives P ] x g = A ] x g $ Q ] x g + R 2 ] x g .
669
EXAMPLES
1. (a) Divide P (x) = 3x 4 - x 3 + 7x 2 - 2x + 3 by x - 2.
(b) Hence write P (x) in the form P (x) = A (x) Q (x) + R (x) .
(c) Show that P (2) is equal to the remainder.
Solution
(a) Step 1: Divide the leading term by x.
i.e. 3x 4 ' x = 3x 3
3x 3
x - 2 g 3 x 4 - x 3 + 7x 2 - 2x + 3
x - 2 is called the divisor.
Step 2: Multiply 3x 3 by (x - 2) and find the remainder by

subtraction.
i.e. 3x 3 (x - 2) = 3x 4 - 6x 3
3x 3
x - 2 g 3x 4 - x 3 + 7x 2 - 2x + 3
3x 4 - 6x 3
5x 3
Step 3: Bring down the 7x 2 and divide 5x 3 by x.
3x 3 + 5x 2
x - 2 g 3x 4 - x 3 + 7x 2 - 2 x + 3
3x 4 - 6x 3
5x 3 + 7x 2
Step 4: Multiply 5x 2 by (x - 2) and find the remainder by
subtraction.
i.e. 5x 2 ] x - 2 g = 5x 3 - 10x 2
3x 3 + 5x 2
x - 2 g 3x 4 - x 3 + 7 x 2 - 2 x + 3
3x 4 - 6x 3
5x 3 + 7x 2
5x 3 - 10x 2
17x 2
CONTINUED
670
The quotient is
3
2
3x + 5x + 17x + 32.
Continuing this way until we have finished, we will have

3x 3 + 5x 2 + 17x + 32
x - 2 g 3x 4 - x 3 + 7 x 2 - 2 x + 3
3x 4 - 6x 3
5x 3 + 7x 2
5x 3 - 10x 2
17x 2 - 2x
17x 2 - 34x
32x + 3
32x - 64
67
The remainder is 67.
(b) This means that (3x 4 - x 3 + 7x 2 - 2x + 3) ' (x - 2)

= (3x 3 + 5x 2 + 17x + 32), remainder 67
3x 4 - x 3 + 7x 2 - 2x + 3
67
= 3x 3 + 5x 2 + 17x + 32 +
x-2
x-2
or 3x 4 - x 3 + 7x 2 - 2x + 3 = (x - 2) (3x 3 + 5x 2 + 17x + 32) + 67
i.e.
i.e. P (x) = A (x) Q (x) + R (x) where A (x) is the divisor, Q (x) is the quotient
and R (x) is the remainder.
(c) P (2) = 3 (2) 4 - 2 3 + 7 (2) 2 - 2 (2) + 3
= 48 - 8 + 28 - 4 + 3
= 67
` P (2) is equal to the remainder.
2. Divide x 3 - 3x 2 + x + 4 by x 2 - x.
Solution
Check this is true by

expanding and
simplifying.
x-2
x 2 - x g x 3 - 3x 2 + x + 4
x3 - x2
- 2x 2 + x
- 2x 2 + 2x
-x + 4
This means that
(x 3 - 3x 2 + x + 4) ' (x 2 - x) = (x - 2), remainder - x + 4
x 3 - 3x 2 + x + 4
-x + 4
=x-2+ 2
i.e.
2
x -x
x -x
or
x 3 - 3x 2 + x + 4 = (x - 2) (x 2 - x) + (- x + 4)
671
3. Divide x 5 + x 3 + 5x 2 - 6x + 15 by x 2 + 3.
Solution
x 3 - 2x + 5
x + 3 g x 5 + x 3 + 5x 2 - 6 x +
x 5 + 3x 3
- 2x 3 + 5x 2 - 6x
- 2x 3
- 6x
2
5x
+
2
5x
+
2
15
15
15
0
R (x) = 0, so there is no
remainder.
This means that

(x 5 + x 3 + 5x 2 - 6x + 15) ' (x 2 + 3) = (x 3 - 2x + 5)
x 5 + x 3 + 5x 2 - 6x + 15
i.e.
= x 3 - 2x + 5
x2 + 3
or
x 5 + x 3 + 5x 2 - 6x + 15 = (x 3 - 2x + 5) (x 2 + 3)
12.2
Exercises
Divide the following polynomials

and put them in the form
P (x) = A (x) Q (x) + R (x) .
11. (6x 2 - 3x + 1) ' (3x - 2)

12. (x 4 - 2x 3 - x 2 - 2) ' (x 2 - x)
13. (3x 5 - 2x 4 - 3x 3 + x 2 - x - 1) '
(x + 2 )
1.
(3x 2 + 2x + 5) ' (x + 4)
2.
(x 2 - 7x + 4) ' (x - 1)
3.
(x + x + 2 x - 1 ) ' (x - 3 )
4.
(4x + 2x - 3) ' (2x + 3)
5.
(x - 5x + x + 2) ' (x + 3x)
6.
(x + x - x - 3) ' (x - 2)
7.
(5x - 2x + 3x + 1) ' (x + x)
8.
(x - x - 2x + x - 3) ' (x + 4)
19. (x 4 - 2x 3 + 4x 2 + 2x + 5) '
(x 2 + 2x - 1)
9.
(2x 4 - 5x 3 + 2x 2 + 2x - 5) '
(x 2 - 2x)
20. (3x 5 - 2x 3 + x - 1) ' (x + 1)
10. (4x 3 - 2x 2 + 6x - 1) ' (2x + 1)
14. (x 2 + 5x - 2) ' (x + 1)
15. (x 4 - 2x 2 + 5x + 4) ' (x - 3)
16. (2x 4 - x 3 + 5) ' (x 2 - 2x)
17. (x 3 - 3x 2 + 3x - 1) ' (x 2 + 5)
18. (2x 3 + 4x 2 - x + 8) ' (x 2 + 3x + 2)
Check this by expanding

and simplifying.
672
Remainder and Factor Theorems

Dividing polynomials helps us to factorise them, which in turn makes
sketching their graphs easier.
There are two theorems that will also help us to work with polynomials.
Remainder theorem
If a polynomial P(x) is divided by x - a, then the remainder is P(a)
Proof
The degree of R(x) is less than
the degree of A(x).
P ] x g = A ] x g $ Q ] x g + R ] x g where A ] x g = x - a
P ]x g = ]x - a gQ ]x g + R ]x g
The degree of A(x) is 1, so the degree of R(x) must be 0.
So
R ] x g = k where k is a constant
`
P ]x g = ]x - a gQ ]x g + k
Substituting
x = a:
P ]a g = ]a - a gQ ]a g + k
= 0 $ Q ]x g + k
=k
So P ] a g is the remainder.
EXAMPLES
1. Find the remainder when 3x 4 - 2x 2 + 5x + 1 is divided by x - 2.
Solution
The remainder when P(x) is divided by x - a is P(a).
The remainder when P(x) is divided by x - 2 is P(2).
P ] 2 g = 3 ] 2 g4 - 2 ] 2 g2 + 5 ] 2 g + 1
= 51
So the remainder is 51.
2. Evaluate m if the remainder is 4 when dividing 2x 4 + mx + 5 by x + 3.
Solution
x + 3 = x - (- 3) .
The remainder when P(x) is divided by x + 3 is P ^ -3 h .

So
P ] -3 g = 4
4
2 ] - 3 g + m ] -3 g + 5 = 4
162 - 3m + 5 = 4
167 - 3m = 4
167 = 3m + 4
163 = 3m
54
1
= m.
3
Factor theorem
The factor theorem is a direct result of the remainder theorem.
For a polynomial P(x), if P ] a g = 0 then x - a is a factor of the polynomial.
Proof
P ] x g = A ] x g $ Q ] x g + R ] x g where A ] x g = x - a
P ]x g = ]x - a gQ ]x g + R ]x g
The remainder when P(x) is divided by x - a is P(a).
So P ] x g = ] x - a g Q ] x g + P ] a g
But if P ] a g = 0:
P ]x g = ]x - a gQ ]x g + 0
= ]x - a gQ ]x g
So x - a is a factor of P(x).
The converse is also true:
For a polynomial P(x), if x - a is a factor of the polynomial, then P ] a g = 0
Proof
If x - a is a factor of P(x), then we can write:
P ]x g = ]x - a gQ ]x g
This means that when P(x) is divided by x - a, the quotient is Q(x) and there is
no remainder.
So P ] a g = 0
673
674
EXAMPLE
(a) Show that x - 1 is a factor of P ] x g = x 3 - 7x 2 + 8x - 2.
(b) Divide P(x) by x - 1 and write P(x) in the form P ] x g = ] x - 1 g Q ] x g.
Solution
(a) The remainder when dividing the polynomial by x - 1 is P(1)
P ] 1 g = 1 3 - 7 ] 1 g2 + 8 ] 1 g - 2
=0
So x - 1 is a factor of P(x).
(b)
x 2 - 6x +
x - 1 g x 3 - 7x 2 +
x3 - x2
- 6x 2 +
- 6x 2 +
Notice that x 2 - 6x + 2
wont factorise.
2
8x - 2
8x
6x
2x - 2
2x - 2
0
3
2
]
g
So P x = x - 7x + 8x - 2
= ] x - 1 g ^ x 2 - 6x + 2 h
Further properties of a polynomial

Some properties of polynomials come from the remainder and factor
theorems.
If polynomial P(x) has k distinct zeros a 1, a 2, a 3, ... a k, then

(x - a 1) (x - a 2) (x - a 3) ... (x - a k) is a factor of P(x)
Proof
If a1 is a zero of P(x) then (x - a 1) is a factor of P(x).
Similarly, if a k is a zero of P(x) then (x - a k) is a factor of P(x).
` P ] x g = (x - a 1) (x - a 2) (x - a 3) ... (x - a k) g ] x g
So (x - a 1) (x - a 2) (x - a 3) ... (x - a k) is a factor of P ] x g .
If polynomial P(x) has degree n and n distinct zeros a 1, a 2, a 3, ... a n,

then P ] x g = p n (x - a 1) (x - a 2) (x - a 3) ... (x - a n)
Proof
Since a 1, a 2, a 3, ... a n are zeros of P(x), (x - a 1) (x - a 2) (x - a 3) ... (x - a n) is a
factor of the polynomial.
So P ] x g = (x - a 1) (x - a 2) (x - a 3) ... (x - a n)Q(x)
But (x - a 1) (x - a 2) (x - a 3) ... (x - a n) has degree n and P(x) has degree n so
Q(x) must be a constant.
` P ] x g = p n (x - a 1) (x - a 2) (x - a 3) ... (x - a n)
A polynomial of degree n cannot have more than n distinct real zeros.
Proof
P(x) has degree n
So P ] x g = p n x n + p n - 1 x n - 1 + p n - 2 x n - 2 + f + p 2 x 2 + p 1 x + p 0 where p n ! 0
Suppose P(x) has more than n distinct zeros, say n + 1
Then (x - a 1) (x - a 2) (x - a 3) ... (x - a n + 1) is a factor of P(x).
So P ] x g = (x - a 1) (x - a 2) (x - a 3) ... (x - a n + 1) Q ] x g.
But this gives P(x) at least degree n + 1, and P(x) only has degree n.
So the polynomial cannot have more than n distinct real zeros.
This also means that the polynomial equation cannot have more than n real roots.
EXAMPLE
If a polynomial has degree 2, show that it cannot have 3 zeros.
Solution
Let P ] x g = p 2 x 2 + p 1 x + p 0 where p 2 ! 0
Assume P(x) has 3 zeros, a1, a2 and a3
Then _ x - a 1 i _ x - a 2 i _ x - a 3 i is a factor of the polynomial.
` P (x) = (x - a 1) (x - a 2) (x - a 3) Q (x)
But this polynomial has degree 3 and P(x) only has degree 2.
So P(x) cannot have 3 zeros.
A polynomial of degree n with more than n distinct real zeros is the

zero polynomial P ] x g = 0 ( p 0 = p 1 = p 2 = ... = p n = 0)
Proof
Let P(x) be a polynomial of degree n with zeros a 1, a 2, a 3, ... a n
Then P ] x g = (x - a 1) (x - a 2) (x - a 3) ... (x - a n) k
675
676
Suppose P(x) has another distinct zero a n + 1

Then P _ a n + 1 i = 0
`(a n + 1 - a 1) (a n + 1 - a 2) (a n + 1 - a 3) ... (a n + 1 - a n) k = 0
But a n + 1 ! a 1, a 2, a 3, ... a n since all zeros are distinct.
So k = 0
`P ] x g = 0
If two polynomials of degree n are equal for more than n distinct values
of x, then the coefficients of like powers of x are equal.
That is, if a 0 + a 1 x + a 2 x 2 + ... + a n x n / b 0 + b 1 x + b 2 x 2 + ... + b n x n
then a 0 = b 0, a 1 = b 1, a 2 = b 2, ... a n = b n
Proof
Let A ] x g = a 0 + a 1 x + a 2 x 2 + ... + a n x n
and B ] x g = b 0 + b 1 x + b 2 x 2 + ... + b n x n
where A ] x g = B ] x g for more than n distinct x values.
Let P ] x g = A ] x g - B ] x g
Then P ] x g = (a 0 - b 0) + (a 1 - b 1) x + (a 2 - b 2) x 2 + ... + (a n - b n) x n
and P(x) has degree n.
If A ] x g = B ] x g for more than n distinct x values, then A ] x g - B ] x g = 0 for more
You learned a special case of this result

in Chapter 10 under quadratic identities.
This is a more general result for all
polynomials.
than n distinct x values.

This means P ] x g = 0 for more than n distinct x values.
This means that P(x) has more than n zeros.
` P(x) is the zero polynomial
P ]xg = 0
(a 0 - b 0) + (a 1 - b 1) x + (a 2 - b 2) x 2 + ... + (a n - b n) x n = 0
So a 0 - b 0 = 0, a 1 - b 1 = 0, a 2 - b 2 = 0, ..., a n - b n = 0
` a 0 = b 0, a 1 = b 1, a 2 = b 2, ..., a n = b n
EXAMPLE
Write x 3 - 2x 2 + 5 in the form ax 3 + b ] x + 3 g2 + c ] x + 3 g + d.
Solution
ax 3 + b ] x + 3 g2 + c (x + 3) + d = ax 3 + b ^ x 2 + 6x + 9 h + c ] x + 3 g + d
= ax 3 + bx 2 + 6bx + 9b + cx + 3c + d
= ax 3 + bx 2 + ] 6b + c g x + 9b + 3c + d
For
x 3 - 2x 2 + 5 / ax 3 + b ] x + 3 g2 + c ] x + 3 g + d
677
]1 g
a=1
b = -2
6b + c = 0
9b + 3c + d = 5
]2 g
]3 g
]4 g
Substitute (2) into (3):

6 ] -2 g + c = 0
- 12 + c = 0
c = 12
Substitute b = - 2 and c = 12 into (4):
9 ] - 2 g + 3 ] 12 g + d
-18 + 36 + d
18 + d
d
=5
=5
=5
= -13
` x 3 - 2x 2 + 5 / x 3 - 2 ] x + 3 g2 + 12 ] x + 3 g - 13.
If x - a is a factor of polynomial P(x), then a is a factor of the constant

term of the polynomial.
Proof
Let P ] x g = p n x n + p n - 1 x n - 1 + p n - 2 x n - 2 + f + p 2 x 2 + p 1 x + p 0 where p n ! 0
If x - a is a factor of P(x) we can write
P ] x g = ] x - a g Q ] x g where Q(x) has degree n - 1.
P ] x g = ] x - a g _ q n - 1 x n - 1 + q n - 2 x n - 2 + g + q 2 x 2 + q 1 x + q 0 i where q n - 1 ! 0
= xq n - 1 x n - 1 + xq n - 2 x n - 2 + g + xq 1 x + xq 0
- aq n - 1 x n - 1 - aq n - 2 x n - 2 - g - aq 2 x 2 - aq 1 x - aq 0
= q n - 1 x n + q n - 2 x n - 1 + g + q 1 x 2 + q 0 x - aq n - 1 x n - 1 - aq n - 2 x n - 2 - g
- aq 2 x 2 - aq 1 x - aq 0
= q n - 1 x n + _ q n - 2 - a i x n - 1 + g + _ q 1 - a i x 2 + _ q 0 - a i x - aq 0
` p 0 = - aq 0
So a is a factor of p0.
We already use this when

factorising a trinomial.
This is a more general
result for all polynomials.
678
EXAMPLE
Factorise x 2 + 2x - 15.
Solution
Factors of -15 are - 3# 5, 3# - 5, -1#15, 1# -15.
We choose - 3# 5 since - 3x + 5x = 2x, the middle term.
So x 2 + 2x - 15 = ] x - 3 g ] x + 5 g.
To factorise polynomials in general, we also look for factors of the constant term.
Class Investigation
Why are factors of the polynomial factors of the constant term? Use the
knowledge you have of trinomials to help you in your discussion.
EXAMPLES
1. Find all factors of f (x) = x 3 + 3x 2 - 4x - 12.
Solution
Try factors of - 12 (i.e. !1, !2, !3, !4, !6, !12) .
e.g. f (1) = 1 3 + 3 (1) 2 - 4 (1) - 12
= -12
!0
` x - 1 is not a factor of f (x)
x - 2 is called a linear
factor as it has degree 1.
f (2) = 2 3 + 3 (2) 2 - 4 (2) - 12

=0
Since f (2) = 0, the remainder when f (x) is divided by x - 2 is 0.
` x - 2 is a factor of f (x) .
We divide f (x) by x - 2 to nd other factors:
x 2 + 5x + 6
x - 2 g x 3 + 3x 2 - 4x - 12
x 3 - 2x 2
5x 2 - 4x
5x 2 - 10x
6x - 12
6x - 12
0
679
` f (x) = (x - 2) (x 2 + 5x + 6)
= (x - 2) (x + 2) (x + 3)
2. Find all factors of P (x) = x 3 + 3x 2 + 5x + 15.
Solution
Try factors of 15 (i.e.!1, !3, !5, !15) .
e.g. P (- 3) = (- 3) 3 + 3 (- 3) 2 + 5 (- 3) + 15
=0
` x + 3 is a factor of f (x)
We divide P (x) by x + 3 to find other factors:
+5
x2
x + 3 g x + 3x + 5x + 15
x 3 + 3x 2
0 + 5x + 15
5x + 15
3
x 2 + 5 will not factorise for

any real x.
` P (x) = (x + 3) (x + 5)
2
12.3 Exercises
1.
Use the remainder theorem

to find the remainder in each
question.
(a) (x 3 - 2x 2 + x + 5) ' (x - 4)
(c) the remainder is 0 when

2x 5 + 7x 2 + 1 + k is divided by
x + 6.
(d) 2x 4 - kx 3 + 3x 2 + x is divisible
by x - 3.
(e) the remainder is 25 when
2x 4 - 3x 2 + 5 is divided by x - k.
(b) (x 2 + 5x + 3) ' (x + 2)
(c) (2x 3 - 4x - 1) ' (x + 3)
(d) (3x 5 + 2x 2 - x + 4) ' (x - 5)
(e) (5x 3 + 2x 2 + 2x - 9) ' (x - 1)
3.
(a) Find the remainder when

f (x) = x 3 - 4x 2 + x + 6 is divided
by x - 2.
(b) Is x - 2 a factor of f (x) ?
(c) Divide x 3 - 4x 2 + x + 6
by x - 2.
(d) Factorise f (x) fully and write
f (x) as a product of its factors.
4.
(a) Show that x + 3 is a factor of

P (x) = x 4 + 3x 3 - 9x 2 - 27x.
(b) Divide P (x) by x + 3 and
write P (x) as a product of its
factors.
(f) (x - x + 3x - x - 1) ' (x + 2)
4
(g) (2x 2 + 7x - 2) ' (x + 7)

(h) (x 7 + 5x 3 - 1) ' (x - 3)
(i) (2x 6 - 3x 2 + x + 4) ' (x + 5)
(j) (3x 4 - x 3 - x 2 - x - 7) ' (x + 1)
2.
Find the value of k if

(a) the remainder is 3 when
5x 2 - 10x + k is divided by x - 1.
(b) the remainder is - 4 when
x 3 - (k - 1) x 2 + 5kx + 4 is divided
by x + 2.
680
5.
The remainder is 5 when

P (x) = ax 3 - 4bx 2 + x - 4
is divided by x - 3 and the
remainder is 2 when P (x) is
divided by x + 1. Find the values
of a and b.
6.
When f (x) = ax 2 - 3x + 1 and

g (x) = x 3 - 3x 2 + 2 are divided
by x + 1 they leave the same
remainder. Find the value of a.
7.
(a) Show that x - 3 is not a factor

of P (x) = x 5 - 2x 4 + 7x 2 - 3x + 5.
(b) Find a value of k such
that x - 3 is a factor of
Q (x) = 2x 3 - 5x + k.
Linear factors are in the

form x - a.
8.
9.
The polynomial
P (x) = x 3 + ax 2 + bx + 2 has
factors x + 1 and x - 2. Find the
values of a and b.
(a) The remainder, when
f (x) = ax 4 + bx 3 + 15x 2 + 9x + 2
is divided by x - 2, is 216, and
x + 1 is a factor of f (x) . Find a
and b.
(b) Divide f (x) by x + 1 and
write the polynomial in the form
f (x) = (x + 1) g (x) .
(c) Show that x + 1 is a factor
of g (x) .
(d) Write f (x) as a product of its
factors.
10. Write each polynomial as a

product of its factors.
(a) x 2 - 2x - 8
(b) x 3 + x 2 - 2x
(c) x 3 + x 2 - 10x + 8
(d) x 3 + 4x 2 - 11x - 30
(e) x 3 - 11x 2 + 31x - 21
(f) x - 12x + 17x + 90
3
(g) x 3 - 7x 2 + 16x - 12
(h) x 4 + 6x 3 + 9x 2 + 4x
(i) x 3 + 3x 2 - 4
(j) x 3 - 7x - 6
11. (a) Write P (x) = x 3 - 7x + 6 as a

(b) What are the zeros of P (x) ?
(c) Is (x - 2) (x + 3) a factor
of P (x) ?
12. If
f (x) = x 4 + 10x 3 + 23x 2 - 34x - 120
has zeros - 5 and 2
(a) show (x + 5) (x - 2) is a factor
of f (x)
(b) write f (x) as a product of its
linear factors.
13. If
P (x) = x 4 + 3x 3 - 13x 2 - 51x - 36
has zeros - 3 and 4, write P (x) as
a product of its linear factors.
14. (a) Show that
P (x) = x 3 - 3x 2 - 34x + 120 has
zeros - 6 and 5.
(b) Write P (x) as a product of its
linear factors.
15. (a) Write the polynomial
P ] u g = u 3 - 4u 2 + 5u - 2 as a
(b) Hence or otherwise, solve
] x - 1 g3 - 4 ] x - 1 g2 + 5 ] x - 1 g - 2 = 0.
16. (a) Write the polynomial
f ^ p h = p 3 - 2p 2 - 5p + 6 as a
] 2x + 1 g3 - 2 ] 2x + 1 g2 - 5 ] 2x + 1 g + 6 = 0.
17. (a) Write P ] k g = 2k 3 + 3k 2 - 1 as
a product of its factors.
2 sin 3 x + 3 sin 2 x - 1 = 0 for
0c # x # 360c.
18. (a) Write
f ] u g = u 3 - 13u 2 + 39u - 27 as a
3 3x - 13.3 2x + 39.3 x - 27 = 0.
19. Solve
] x + 4 g4 - ] x + 4 g3 - 2 ] x + 4 g2 = 0.
20. Solve 2 cos 3 i - cos 2 i - cos i = 0

for 0c# x # 360c.
21. Evaluate a, b, c and d if
(a) x 3 + 3x 2 - 2x + 1 /
ax 3 + b ] x - 1 g2 + cx + d
(b) x 3 - x 2 + 4x /
ax 3 + b ] x + 2 g2 + c ] x + 2 g + d
(c) 2x 3 - x + 7 /
ax 3 + b ] x + 1 g2 + c ] x + 1 g + d + 2
(d) x 3 + x 2 + 5x - 3 /
ax 3 + b ] x - 3 g2 + cx + d
(e) 4x 3 - x + 3 /
] a + 1 g x 3 + b ] x + 4 g2 +
c ]x + 4 g + d - 1
(f) x 3 + x 2 - 8x - 6 /
ax 3 + b ] x - 2 g2 + cx + d - 3
(g) 3x 3 - 2x 2 + x /
] a - 2 g x 3 + b ] x - 5 g2 +
c ]x - 5g + d - 2
(h) - x 3 + x 2 - 4x - 2 /
a ] x + 1 g3 + bx 2 + cx + d
(i) - 2x 3 + 3x 2 - 1 /
2ax 3 + b ] x - 1 g2 + cx + d
(j) - x 3 - 4x 2 + x + 3 /
a ] x - 2 g3 + b ] x - 2 g2 +
c ]x - 2 g + d + 1
22. A monic polynomial of degree
3 has zeros - 3, 0 and 4. Find the
polynomial.
23. Polynomial
P ] x g = ax 3 - bx 2 + cx - 8 has
zeros 2 and - 1 and P ] 3 g = 28.
Evaluate a, b and c.
24. A polynomial with leading term
2x4 has zeros - 2, 0, 1 and 3. Find
the polynomial.
25. Show that a polynomial of degree
3 cannot have 4 zeros.
Graph of a Polynomial
We can use the graphing techniques that you have learned to sketch the graph
of a polynomial.
Using intercepts
Finding the zeros of a polynomial or the roots of the polynomial equation
helps us to sketch its graph.
EXAMPLES
1. (a) Write the polynomial P ] x g = x 3 + x 2 - 6x as a product of its factors.
(b) Sketch the graph of the polynomial.
Solution
(a) P ] x g = x 3 + x 2 - 6x
= x ^ x2 + x - 6 h
= x ]x + 3 g]x - 2 g
(b) For the graph of P ] x g = x 3 + x 2 - 6x
For x-intercepts: y = 0
CONTINUED
681
682
0 = x 3 + x 2 - 6x
= x ]x + 3 g]x - 2 g
x = 0,
x + 3 = 0, x - 2 = 0
x = - 3,
x=2
So x-intercepts are 0, - 3 and 2.
For y-intercepts: x = 0
P ] 0 g = 0 3 + ] 0 g2 - 6 ] 0 g
=0
So y-intercept is 0.
y
4
3
2
1
-4
-3
-2
-1 0
-1
x
1
-2
-3
-4
We look at which parts of the graphs are above and which are below the
x-axis between the x-intercepts.
Test x 1 - 3, say x = - 4:
P ] x g = x 3 + x 2 - 6x
= x ]x + 3 g]x - 2 g
P ]-4 g = -4 ]-4 + 3 g]-4 - 2 g
= -4 ]-1 g]-6 g
= - 24
10
So the curve is below the x-axis.
Test - 3 1 x 1 0, say x = - 1:
P ] - 1 g = - 1 ] - 1 + 3 g ] -1 - 2 g
= -1 ]2 g]-3 g
=6
20
So the curve is above the x-axis.
683
Test 0 1 x1 2, say x = 1:
P ]1g = 1]1 + 3g]1 - 2g
= 1]4g]-1g
= -4
10
Test x 2 2, say x = 3:
P ]3g = 3]3 + 3g]3 - 2g
= 3]6g]1g
= 18
20
We can sketch the polynomial as shown.
y
4
3
2
1
-4
-3
-2
-1 0
-1
x
1
-2
-3
-4
2. (a) Write the polynomial P ] x g = x 3 - x 2 - 5x - 3 as a product of its

factors.
(b) Sketch the graph of the polynomial.
Solution
(a) Factors of - 3 are ! 1 and ! 3.
P ] -1 g = ] -1 g3 - ] -1 g2 - 5 ] -1 g - 3
=0
CONTINUED
Later on, in a class investigation

in this chapter you will learn
how to make the graph
more accurate by finding the
maximum and minimum points.
This is a topic in the HSC Course.
684
So x + 1 is a factor of the polynomial.

x 2 - 2x - 3
x + 1 g x 3 - x 2 - 5x - 3
x3 + x2
-2x2 - 5x
-2x2 - 2x
-3x - 3
-3x - 3
0
P ]x g = ]x +
= ]x +
= ]x +
1 g ^ x 2 - 2x - 3 h
1 g]x - 3 g]x + 1 g
1 g2 ] x - 3 g
(b) For the graph of P ] x g = x 3 - x 2 - 5x - 3

0 = x 3 - x 2 - 5x - 3
= ] x + 1 g2 ] x - 3 g
] x + 1 g2 = 0,
x-3=0
x+1=0
x=3
x = -1
So x-intercepts are - 1 and 3.
For y-intercepts: x = 0
y
4
3
2
1
-4
-3
-2
-1 0
-1
P ] 0 g = 0 3 - ] 0 g2 - 5 ] 0 g - 3
= -3
-2
So y-intercept is - 3.
-4
x
1
-3
We look at which parts of the graphs are above and which are below the
x-axis between the x-intercepts.
Test x 1 -1, say x = - 2:
P ] x g = x 3 - 3x 2 - x + 3
= ] x + 1 g2 ] x - 3 g
P ] - 2 g = ] - 2 + 1 g2 ] - 2 - 3 g
= ] - 1 g2 ] - 5 g
= -5
10
Test -11 x 1 3, say x = 0:
P ] 0 g = ] 0 + 1 g2 ] 0 - 3 g
= ] 1 g2 ] - 3 g
= -3
10

Test x 2 3, say x = 4:
P ] 4 g = ] 4 + 1 g2 ] 4 - 3 g
= ] 5 g2 ] 1 g
= 25
20
We can sketch the polynomial as shown.
y
4
3
2
1
-4
-3
-2
-1 0
-1
-2
-3
-4
-5
-6
-7
12.4 Exercises
1.
Sketch the graph of each

polynomial by nding its zeros
and showing the x- and
y-intercepts.
(a) f ] x g = ] x + 1 g ] x - 2 g ] x - 3 g
(b) P ] x g = x ] x + 4 g ] x - 2 g
2.
(i) Write each polynomial as a

product of its factors
(ii) Sketch the graph of the
polynomial
(a) P ] x g = x 3 - 2x 2 - 8x
(c) p ] x g = - x ] x - 1 g ] x - 3 g
(d) f ] x g = x ] x + 2 g2
(b) f ] x g = - x 3 - 4x 2 + 5x
(c) P ] x g = x 4 + 3x 3 + 2x 2
(d) A ] x g = 2x 3 + x 2 - 15x
(e) g ] x g = ] 5 - x g ] x + 2 g ] x + 5 g
(e) P ] x g = - x 4 + 2x 3 + 3x 2
685
686
3.
4.
5.
(a) Find the x-intercepts of the

polynomial P ] x g = x 4 + 3x 3 - 4x.
(b) Sketch the graph of the
polynomial.
(a) Show that x - 2 is a factor of
P ] x g = x 3 - 3x 2 - 4x + 12.
(b) Write P(x) as a product of its
factors.
(c) Sketch the graph of the
polynomial.
(a) P ] x g = x 3 + 3x 2 - 10x - 24
(b) P ] x g = x 3 + x 2 - 9x - 9
(c) P ] x g = 12 - 19x + 8x 2 - x 3
(d) P ] x g = x 3 - 13x + 12
(e) P ] x g = - x 3 + 2x 2 + 9x - 18
(f) P ] x g = x 3 + 2x 2 - 4x - 8
(g) P ] x g = x 3 - 5x 2 + 8x - 4
(h) P ] x g = x 3 + x 2 - 5x + 3
(i) f (x) = 16x + 12x 2 - x 4
(j) P ] x g = x 4 - 2x 2 + 1
Sketch the graph of each

polynomial, showing all x- and
y-intercepts.
Class Investigation
The graphs in the examples above are not very accurate, as we dont know
where they turn around. We can use calculus to help find these points.
You will look at the
applications of calculus in
sketching graphs in the HSC
Course.
You used the axis of symmetry to find the minimum and maximum
values of quadratic functions in Chapter 10. You can also use calculus to
find the minimum or maximum turning points of functions.
You looked at the gradient

of tangents to a curve in
Chapter 8.
Notice that the graph below has both a maximum and minimum turning
point. We can find these by looking at the gradient of the tangents
dy
around the curve, or
.
dx
y
Maximum turning point
Minimum turning point
Notice that at both these turning points,
dy
dx
= 0.
We can also examine each type of turning point more closely.

Maximum turning point:
The maximum turning point has a zero gradient at the point itself but
notice that it has a positive gradient on the left-hand side and a negative
gradient on the right-hand side.
dy
dy
So
2 0 on the LHS and
1 0 on the RHS.
dx
dx
Minimum turning point:
The minimum turning point has a zero gradient at the point itself but it
has a negative gradient on the left-hand side and a positive gradient on
the right-hand side.
dy
dy
So
1 0 on the LHS and
2 0 on the RHS.
dx
dx
CONTINUED
687
688
There is also another type of point that you see in graphs such as f ] x g = x 3.
This is called a point of inexion and has
dy
dx
= 0.
However, the gradient has the same sign on both the LHS and RHS.
These three types of points are called stationary points.
We can use them to sketch the graph of a polynomial. Here is an example.
Sketch the polynomial P ] x g = 2x 3 + 3x 2 - 12x - 7 showing any stationary
points.
dy
dx
= 6x 2 + 6x - 12
For stationary points
dy
dx
= 0:
6x 2 + 6x - 12 = 0
6 ^ x2 + x - 2 h = 0
6 ]x - 1 g]x + 2 g = 0
x - 1 = 0, x + 2 = 0
x = 1,
x = -2
So there are two stationary points when x = 1, - 2.
] 1 g When x = 1
P ] 1 g = 2 ] 1 g3 + 3 ] 1 g2 - 12 ] 1 g - 7
= -14
So there is a stationary point at ^ 1, -14 h.
We can check the gradient on the LHS and RHS of this point to determine
if it is a maximum or minimum turning point.
When x = 0
dy
= 6 (0) 2 + 6 (0) - 12
dx
= -12
When x = 2
dy
= 6 (2) 2 + 6 (2) - 12
dx
= 24
x
dy
dx
Since
dy
-12
24
1 0 on the LHS and
dx
turning point.
dy
dx
2 0 on the RHS, ^ 1, -14 h is a minimum
] 2 g When x = - 2
P ] - 2 g = 2 ] - 2 g3 + 3 ] - 2 g2 - 12 ] - 2 g - 7
= 13
So there is a stationary point at ^ - 2, 13 h.
Check the gradient on the LHS and RHS of this point.
When x = - 3
dy
= 6 (- 3) 2 + 6 (- 3) - 12
dx
= 24
When x = -1
dy
= 6 (-1) 2 + 6 (-1) - 12
dx
= -12
x
dy
dx
Since
dy
dx
-3
-2
-1
24
-2
2 0 on the LHS and
dy
dx
1 0 on the RHS, ^ - 2, 13 h is a maximum
turning point.
Now we look for intercepts.
0 = 2x 3 + 3x 2 - 12x - 7
The expression 2x 3 + 3x 2 - 12x - 7 will not factorise so we cannot find
the x-intercepts.
For y-intercept: x = 0
P ] x g = 2 ] 0 g3 + 3 ] 0 g2 - 12 ] 0 g - 7
= -7
So the y-intercept is - 7.
CONTINUED
Factors of - 7 are !1
and !7 and none of
these factors will satisfy
the polynomial equation.
689
690
We sketch the polynomial using the stationary points and y-intercept.

y
(-2, 13)
x
-7
(1, -14)
Can you sketch the following polynomials using calculus to find their
stationary points?
1. P ] x g = x 2 + 6x - 3
2. P ] x g = - x 2 + 4x + 1
3. p ] x g = x 3 - 5
4. f ] x g = x 4 + 2
5. g ] x g = 2x 3 + 3x 2 - 1
6. P ] x g = 2x 3 - 21x 2 + 72x - 12
7. f ] x g = - 2x 3 + 9x 2 - 12x + 4
8. P ] x g = x 3 - 3x 2 + 3x - 5
9. A ] x g = x 4 + 8x 3 - 18x 2 - 7
10. Q ] x g = - 3x 4 + 20x 3 - 48x 2 + 48x - 3
You may have noticed some of these properties while sketching the graphs of
polynomials.
Limiting behaviour of polynomials

The limiting behaviour of a function describes what happens to the function as
x " !3.
For very large x , P (x) . p n x n
Investigation
Use a graphics calculator or graphing computer software to explore the
behaviour of polynomials as x becomes large (both negative and positive
values).
For example, sketch f ] x g = 2x 5 + 3x 2 - 7x - 1 and f ] x g = 2x 5 together.
What do you notice at both ends of the graphs where x is large? Zoom
out on these graphs and watch the graph of the polynomial and the
graph of the leading term come together.
Try sketching other polynomials along with their leading term as
different graphs. Do you find the same results?
So the leading term shows us what its limiting behaviour will be.
If the degree of a polynomial P(x) is even and the leading coefficient is
positive, then the polynomial will be positive as x becomes large.
This means that for any polynomial with a positive leading coefficient
and even degree, P ] x g " 3 as x " !3.
On the graph, both ends of the graph will go up as shown by the
examples below.
y
All positive or negative

values of x to an even
power will always be
positive.
691
692
If the degree of a polynomial P(x) is even and the leading coefficient is

negative, then the polynomial will be negative as x becomes large.
This means that for any polynomial with a negative leading coefficient
and even degree, P ] x g " - 3 as x " ! 3.
On the graph, both ends of the graph will go down as shown by the
examples below.
y
If P(x) is an odd degree polynomial with positive leading coefficient, then

as x becomes a very large positive value, P(x) will also be positive. As x becomes
a very large negative value, P(x) will also be negative.
This means that P ] x g " - 3 as x " - 3 and P ] x g " 3 as x " 3.
On the graph, the end of the graph on the LHS will go down and the end
on the RHS will go up as shown in the examples.
y
693
694
If P(x) is an odd degree polynomial with negative leading coefficient, then

as x becomes a very large positive value, P(x) will be negative. As x becomes
a very large negative value, P(x) will be positive.
This means that P ] x g " 3 as x " - 3 and P ] x g " - 3 as x " 3.
On the graph, the end of the graph on the LHS will go up and the end on
the RHS will go down as shown in the examples.
y
695
696
If P(x) has even degree, the ends of the graph both go the same way.
y
Leading coefficient 2 0
y
If P(x) has odd degree, the ends of the graph both go different ways.
y
A polynomial of odd degree always has at least one real zero.
This comes from the results above. A polynomial with odd degree will go
up at one end and down the other as x becomes large. This means that it must
cross the x-axis at least once.
` the polynomial must have at least one real zero.
At least one maximum or minimum value of P(x) occurs

between any two distinct real zeros.
You can see this on a graph. If there are two distinct real zeros of a
polynomial, then they will show up on the graph as two x-intercepts since the
zeros make P ] x g = 0.
When the graph passes through one x-intercept, say x1, it must turn
around again to pass through the other x-intercept x2 as shown in the
examples below. So there must be at least one maximum or minimum value
between the zeros.
y
x1
x2
x1
x2
x1
x2
Multiple roots
In quadratic functions, you saw that if a quadratic expression is a perfect
square, it has equal roots (and the discriminant is zero).
697
698
EXAMPLE
Solve x 2 - 2x + 1 = 0.
Solution
x 2 - 2x + 1 = 0
]x - 1g]x - 1g = 0
] x - 1 g2 = 0
x - 1 = 0,
x = 1,
x -1= 0
x =1
The solution is x = 1 but we say that there are two equal roots.
If P(x) has two equal roots at x = a then we can write P ] x g = ] x - a g2 Q ] x g

We say that the polynomial has a double root at x = a.
If P ] x g = ] x - a g3 Q ] x g, the polynomial has a triple root at x = a. There are
three equal roots at x = a.
If P ] x g = ] x - a gn Q ] x g, the polynomial has a multiple root at

x=a
It has n equal roots at x = a
EXAMPLES
1. Sketch the graph of f ] x g = ] x + 2 g2.
Solution
This graph is f ] x g = x 2 translated 2 units to the left.
y
5
4
3
2
1
-4 -3 -2 -1 0
-1
x
1
-2
-3
-4
See class investigations
on pages 686690.
Notice that there is a minimum turning point at the root x = - 2.
699
2. Sketch the graph of F ] x g = ] x - 1 g3.
Solution
This is the graph of F ] x g = x 3 translated 1 unit to the right.
y
8
6
4
2
-4
-3
-2
-1 0
-2
x
1
-4
-6
-8
See class investigations
on pages 686690.
Notice that there is a point of inflexion at the root x = 1.
Generally, a graph cuts the x-axis at a single root but touches the x-axis at
a multiple root in a special way.
EXAMPLE
(a) Examine the polynomial P ] x g = ] x + 2 g2 ] x - 1 g close to the roots.
(b) Describe the behaviour of the polynomial as x becomes very large.
(c) Draw a sketch of the polynomial showing its roots.
Solution
(a) P ] x g = ] x + 2 g2 ] x - 1 g has roots when P ] x g = 0.
] x + 2 g2 ] x - 1 g = 0
x + 2 = 0, x - 1 = 0
x = - 2,
x=1
Notice that there is a double root at x = - 2.

Look at the sign of P(x) close to x = 1:
When x = 0.9
P ] 0.9 g = ] 0.9 + 2 g2 ] 0.9 - 1 g
= +#=So the curve is below the x-axis on the LHS.
CONTINUED
700
When x = 1.1
P ] 1.1 g = ] 1.1 + 2 g2 ] 1.1 - 1 g
= +#+
=+
So the curve is above the x-axis on the RHS.
Look at the sign of P(x) close to x = - 2:
When x = - 2.1
P ] - 2.1 g = ] - 2.1 + 2 g2 ] - 2.1 - 1 g
= +#=So the curve is below the x-axis on the LHS.
When x = - 1.9
P ] - 1.9 g = ] - 1.9 + 2 g2 ] - 1.9 - 1 g
= +#=So the curve is below the x-axis on the RHS.
At the single root x = 1, the curve passes through the root from
below the x-axis to above the x-axis.
At the double root x = - 2, the curve touches the x-axis from below
and turns around and continues to be below the x-axis.
(b) Expanding P ] x g = ] x + 2 g2 ] x - 1 g gives x3 as the leading term.
P ] x g = ] x + 2 g2 ] x - 1 g
= ^ x 2 + 4x + 4 h ] x - 1 g
There is no need to expand
the brackets fully as we only
need the leading term.
= x 3 - x 2 + 4x 2 - 4x + 4 x - 4
So the polynomial has degree 3 since the highest power is x3.
Also the leading coefficient is 1.
Since P(x) has odd degree and a positive leading coefficient, as x becomes
a larger positive number, P ] x g "3 and as x becomes a larger negative
number, P ] x g " -3.
(c)
-2
701
Investigation
Use a graphics calculator or graphing computer software to draw graphs
with multiple roots.
(a) Examine values close to the roots.
(b) Look at the relationship between the degree of the polynomial, the
leading coefficient and its graph.
Here are some examples of polynomials but you could choose others to
examine.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
P ]xg = ]x + 1g]x - 3g
P ]xg = ]x + 1g2]x - 3g
P ]xg = -]x + 1g3]x - 3g
P ]xg = -]x + 1g4]x - 3g
P ]xg = ]x + 1g]x - 3g2
P ]xg = ]x + 1g]x - 3g3
P ]xg = -]x + 1g]x - 3g4
P ]xg = -]x + 1g2]x - 3g2
P ]xg = -]x + 1g2]x - 3g3
P ]xg = ]x + 1g3]x - 3g2
Where there is a multiple root, there is always a stationary point

(maximum, minimum or point of inflexion).
This means that
dy
dx
= 0 at that point.
If the root is at x = a, then we can write this as Pl(a) = 0 .
If P ] x g = ] x - a gn Q ] x g has a multiple root at x = a then P (a) = P l(a) = 0

There is a stationary point at x = a:
If n is even, there is a maximum or minimum turning point at x = a
If n is odd, there is a point of inflexion at x = a
Proof
P ] x g = ] x - a gn Q ] x g
P ] a g = ] a - a gn Q ] a g
= 0n $ Q ] x g
=0
See class
investigation on
pages 686690.
702
P l (x) = u lv + v lu
= n (x - a) n - 1 $ 1 $ Q (x) + Ql(x) $ (x - a) n
= n (x - a) n - 1 Q (x) + (x - a) n Ql(x)
P l (a) = n (a - a) n - 1 Q (a) + (a - a) n Ql(a)
= n $ 0 n - 1 $ Q (a) + 0 n $ Ql(a)
=0
` P (a) = P l (a) = 0
EXAMPLES
1. Draw a sketch of P ] x g = - x ] x - 3 g3 .
Solution
Roots of the polynomial equation P ] x g = 0:
- x ] x - 3 g3 = 0
x = 0, x - 3 = 0
x=3
There is a single root at x = 0 so the curve crosses the x-axis at this point.
There is a triple root at x = 3. Since n is odd, there is a point of inflexion
at x = 3.
P ] x g = - x ] x - 3 g3
There is no need to fully expand
the polynomial as we only want
to find the leading term.
= - x ^ x 3 - 9x 2 + 27x - 27 h
= - x4 f
Since x 4 is the leading term, P(x) has degree 4 and the leading coefficient
is negative.
So as x becomes large (both negative and positive) the value of
P ] x g " -3.
y
703
2. A polynomial has a double root at x = 5.

(a) Write an expression for the polynomial.
(b) Prove that P ] 5 g = P l (5) = 0.
Solution
(a) If P(x) has a double root at x = 5, then ] x - 5 g 2 is a factor
So P ] x g = ] x - 5 g 2Q ] x g
(b) P ] x g = ] x - 5 g 2Q ] x g
P ] 5 g = ] 5 - 5 g 2Q ] 5 g
= 02 #Q ] 5 g
=0
To find P l (5), first we differentiate P(x) using the product rule.
P l (x) = u lv + v lu
= 2 (x - 5) 1 $ 1 $ Q (x) + Q l (x) $ (x - 5) 2
= 2 ( x - 5 ) Q ( x ) + ( x - 5 ) 2 Q l (x )
2
P l ( 5 ) = 2 (5 - 5 ) Q ( 5 ) + (5 - 5 ) Q l ( 5 )
2
= 2 # 0 # Q (5 ) + 0 # Q l ( 5 )
=0
3. A monic polynomial has degree 5 and has a double root at a1 and a
triple root at a2. Draw a sketch of the polynomial where a 1 1 a 2 .
Solution
Since P(x) is monic and has degree 5, the leading term is x5.
We could write P ] x g = _ x - a 1 i 2 _ x - a 2 i 3.
Since the polynomial has odd degree and a positive leading coefficient, as
x becomes a positive large value, P ] x g " 3 and as x becomes a negative
large value, P ] x g " - 3.
The double root at x = a 1 gives a maximum or minimum turning point
and the triple root at x = a 2 means a point of inflexion.
Putting all this information together gives the graph below.
y
a1
a2
You learned this

rule in Chapter 4.
704
12.5 Exercises
1.
Find the roots of each polynomial

equation P (x) = 0 and state if
they are multiple roots.
(a) P ] x g = x 2 - 6x + 9
(b) P ] x g = x 3 - 9x 2 + 14x
(c) P ] x g = x 3 - 3x 2
(d) f ] x g = x 3 - 2x 2 - 4x + 8
(e) P ] x g = x 3 - 6x 2 + 12x - 8
(f) A ] x g = x 4 - 4x 3 + 5x 2 - 2x
(g) P ] x g = x 4 - 4x 3 - 2x 2 +
12x + 9
(h) Q ] x g = x 5 - 8x 4 + 16x 3
(i) P ] x g = x 4 + 2x 3 - 12x 2 +
14x - 5
(j) f ] x g = 8x 3 - 36x 2 + 54x - 27
(c)
(d)
2.
For each graph, state if

(i) the leading coefficient is
positive or negative and
(ii) the degree of the polynomial
is even or odd.
(a)
(e)
(b)
(f)
x
x
(g)
(b) If P ] 2 g = 5, write the

expression for the polynomial.
5.
Polynomial
P ] x g = x 3 - 7x 2 + 8x + 16 has a
double root at x = 4.
(a) Show that ] x - 4 g 2 is a factor
of P(x).
(b) Write P(x) as a product of its
factors.
(c) Prove P ] 4 g = P l(4) = 0.
6.
Polynomial
f ] x g = x 4 + 7x 3 + 9x 2 - 27x - 54
has a triple root at x = - 3.
(a) Show that ] x + 3 g3 is a factor
of f(x).
(b) Write f (x) as a product of its
factors.
(c) Prove f ] - 3 g = f l(- 3) = 0 .
7.
A polynomial has a triple root at

x = k and degree n.
(a) Write an expression for the
polynomial.
(b) Prove that P (k) = P l (k) = 0.
8.
Draw an example of a polynomial

with leading term
(a) x3
(b) - 2x 5
(c) 3x2
(d) - x 4
(e) - 2x 3
9.
Draw an example of a polynomial

with a double root at x = 2 and a
leading term of 2x3.
(h)
(i)
(j)
3.
4.
A monic polynomial of degree

2 has a double root at x = - 4.
Write down an expression for the
polynomial P(x). Is this a unique
expression?
A polynomial of degree 3 has a
triple root at x = 1.
(a) Write down an expression for
the polynomial. Is this unique?
10. Draw an example of a polynomial

with a double root at x = -1 and
leading term - x 3.
11. Sketch an example of a
polynomial with a double root at
x = 2 and a leading term of x4.
12. Draw an example of a polynomial
with a double root at x = - 3 and
leading term x 6 .
705
706
13. A polynomial has a triple root at

x = 1 and it has a leading term of
x3. Draw an example of a graph
showing this information.
14. Given a polynomial with a triple
root at x = 0 and leading term
- x 4, sketch a polynomial on
a number plane that fits this
information.
15. If a polynomial has a triple root
at x = - 2 and a leading term of
x8 sketch a polynomial fitting this
information.
16. A polynomial has a triple root
at x = 4 and its leading term is
- 4x 3 . Show this on a number
plane.
18. A polynomial with leading term

- x 8 has a triple root at x = - 2.
Show by a sketch that the
polynomial has at least one other
root in the domain x 2 - 2.
19. A polynomial has a double root
at x = 2 and a double root at
x = - 3. Its leading term is 2x5. By
drawing a sketch, show that the
polynomial has another root in
the domain x 2 2.
20. Show that a polynomial with
leading term - x 3 and a double
root at x = 1 has another root at a
point where x 21.
17. A monic polynomial has degree

3 and a double root at x = -1.
Show on a sketch that the
polynomial has another root in
the domain x 2 -1.
Roots and Coefficients of

Polynomial Equations
In Chapter 10, you studied the relationship between the roots and coefficients
of the quadratic equation. In this section you will revise this and also study
this relationship for cubic and quartic equations.
Quadratic equation
The quadratic equation ax 2 + bx + c = 0 can be written in monic form as
c
b
x2 + a x + a = 0
If the quadratic equation has roots a and b, then the equation can be
written in monic form as
( x - a ) (x - b ) = 0
x 2 - bx - a x + ab = 0
x 2 - (a + b ) x + ab = 0
i.e.
c
b
x 2 + a x + a / x 2 - (a + b) x + ab
This gives the results below:
For the quadratic equation ax 2 + bx + c = 0:

Sum of roots:
b
a + b = -a
Product of roots:
c
ab = a
Cubic equation
The cubic equation ax 3 + bx 2 + cx + d = 0 can be written in monic form as
d
c
b
x 3 + a x 2 + a x + a = 0.
If the cubic equation has roots a, b and c then the equation can be
(x - a ) (x - b ) (x - c) = 0
(x 2 - bx - a x + ab ) (x - c) = 0
x 3 - cx 2 - bx 2 + bcx - a x 2 + acx + abx - abc = 0
x 3 - (a + b + c) x 2 + (ab + bc + ac) x - abc = 0
d
c
b
x 3 + a x 2 + a x + a / x 3 - (a + b + c) x 2 + (ab + bc + ac) x - abc
For the cubic equation ax 3 + bx 2 + cx + d = 0:

Sum of roots 1 at a time:
b
a + b + c = -a
c
ab + ac + bc = a
Product of roots (sum of roots 3 at a time)
d
abc = - a
Quartic equation
The quartic equation ax 4 + bx 3 + cx 2 + dx + e = 0 can be written in monic form
d
c
e
b
as x 4 + a x 3 + a x 2 + a x + a = 0.
707
708
If the quartic equation has roots a, b, c and d then the equation can be
(x - a ) (x - b) (x - c) (x - d) = 0
[x - (a + b + c) x + (ab + bc + ac) x - abc] (x - d) = 0
x - dx - (a + b + c) x 3 + d (a + b + c) x 2 +
(ab + bc + ac) x 2 - d (ab + bc + ac) x - abcx + abcd = 0
x 4 - (a + b + c + d) x 3 + (ad + db + dc + ab + bc + ac) x 2 (abd + bdc + adc + abc) x + abcd = 0
e
b 3 c 2 d
4
` x + ax + ax + ax + a
3
/ x 4 - (a + b + c + d) x 3 + (ad + db + dc + ab + bc + ac) x 2
- (abd + bdc + adc + abc) x + abcd
For the quartic equation ax 4 + bx 3 + cx 2 + dx + e = 0:

b
a + b + c + d = -a
c
ab + ac + ad + bc + bd + cd = a
d
abc + abd + acd + bcd = - a
Product of roots (sum of roots 4 at a time):
e
abcd = a
This pattern extends to polynomials of any degree.
Class Investigation
Can you find results for sums and products of roots for equations of
degree 5, 6 and so on?
EXAMPLES
1. If a, b, c are the roots of 2x 3 - 5x 2 + x - 1 = 0, find
(a ) (a + b + c) 2
(b) (a + 1) ( b + 1) (c + 1)
(c)
1 1 1
+ + .
a b c
Solution
b
a + b + c = -a
=-
(- 5)
2
5
2
c
ab + ac + bc = a
1
=
2
=
d
abc = - a
==
(-1)
2
1
2
5 2
1
(a) (a + b + c) 2 = c m = 6
4
2
(b) (a + 1) ( b + 1) (c + 1)
= (a + 1) ( bc + b + c + 1)
= abc + ab + ac + a + bc + b + c + 1
= abc + (ab + ac + bc) + (a + b + c) + 1
5
1
1
= + + +1
2
2
2
1
=4
2
bc + ac + ab
(c) 1
1
1
+ + =
a
c
b
abc
1
2
=
1
2
=1
CONTINUED
709
710
2. If one root of x 3 - x 2 + 2x - 3 = 0 is 4, find the sum and product of

the other two roots.
Solution
Roots are a, b, c where, say, c = 4.
b
a + b + c = -a
`
a+b+4 =1
a + b = -3
d
abc = - a
ab (4) = 3
3
ab =
4
3. Solve 12x 3 + 32x 3 + 15x - 9 = 0, given that 2 roots are equal.
Solution
Let the roots be a, a and b.
b
a + a + b = -a
`
2a + b = -
32
12
(1)
c
aa + ab + ab = a
15
`
a 2 + 2ab =
12
d
aab = - a
9
`
a2 b =
12
(2)
(3)
From (1):
b=-
32
- 2a
12
Substitute in (2):
32
15
a 2 + 2a c - 2a m =
12
12
32
2
12a + 24a c - 2a m = 15
12
12a 2 - 64a - 48a 2 = 15
0 = 36a 2 + 64a + 15
= (2a + 3) (18a + 5)
2a + 3 = 0
18a + 5 = 0
2a = - 3
18a = - 5
-5
1
a = -1
a=
2
18
(4)
Substitute in (4):
32
1
1
a = -1 : b = - 2 c -1 m
12
2
2
1
=
3
32
5
5
a=- :b=- 2cm
12
18
18
1
= -2
9
Substitute in (3):
9
1
1
1 2 1
a = - 1 , b = : c- 1 m c m =
2
3
2
3
12
3
3
=
4
4
5
5 2
9
1
1
a = - , b = - 2 : cm c- 2 m =
18
9
18
9
12
This is impossible as LHS is negative and RHS is positive.
` the roots are -1
1
1
and
2
3
12.6 Exercises
1.
Given that a and b are the roots

of the equation, find
(i) a + b and
(ii) ab for the following quadratic
equations.
(a) x 2 - 2x + 8 = 0
(b) 3x 2 + 6x - 2 = 0
(c) x 2 + 7x + 1 = 0
(d) 4x 2 - 9x - 12 = 0
(e) 5x 2 + 15x = 0
2.
Find
(i) a + b + c,
(ii) ab + ac + bc, and
(iii) abc, where a, b and c are
the roots of the equation, for the
following cubic equations.
(a) x 3 + x 2 - 2x + 8 = 0
(b) x 3 - 3x 2 + 5x - 2 = 0
(c) 2x 3 - x 2 + 6x + 2 = 0
(d) - x 3 - 3x 2 - 11 = 0
(e) x 3 + 7x - 3 = 0
3.
For the following quartic

equations, where a, b, c and d are
the roots of the equation, find
(i) a + b + c + d,
(ii) ab + ac + ad + bc + bd + cd,
(iii) abc + abd + acd + bcd and
(iv) abcd
(a) x 4 + 2x 3 - x 2 - x + 5 = 0
(b) x 4 - x 3 - 3x 2 + 2x - 7 = 0
(c) - x 4 + x 3 + 3x 2 - 2x + 4 = 0
(d) 2x 4 - 2x 3 - 4x 2 + 3x - 2 = 0
(e) 2x 4 - 12x 3 + 7 = 0
4.
If a and b are the roots of

x 2 - 5x - 5 = 0, find
(a) a + b
(b) ab
1 1
(c)
+
a b
(d) a2 + b2
(e) a3 + b3
711
712
5.
6.
If a , b and c are the roots of

2x 3 + 5x 2 - x - 3 = 0, nd
(a) abc
(b) ab + ac + bc
(c) a + b + c
1 1 1
(d) + +
a b c
(e) (a + 1) ( b + 1) (c + 1)
If a, b, c and d are the roots of
x 4 - 2x 3 + 5x - 3 = 0, nd
(a) abcd
(b) abc + abd + acd + bcd
1 1 1 1
(c)
+ + +
a b c d
7.
One root of x 2 - 3x + k - 2 = 0
is - 4. Find the value of k.
8.
One root of x 3 - 5x 2 - x + 21 = 0
is 3. Find the sum a + b and
the product ab of the other
two roots.
9.
Given P (x) = 2x 3 - 7x 2 + 4x + 1,
if the equation P (x) = 0 has a
root at x = 1, nd the sum and
product of its other roots.
10. Find the value(s) of k if

the quadratic equation
x 2 - (k + 2) x + k + 1 = 0 has
(a) equal roots
(b) one root equal to 5
(c) consecutive roots
(d) one root double the other
(e) reciprocal roots
11. Two roots of
x 3 + mx 2 + 15x - 7 = 0 are equal
and rational. Find m.
12. Two roots of

x 3 + ax 2 + bx - 5 = 0 are equal
to 4 and - 2. Find the values of
a and b.
13. (a) Show that 1 is a
zero of the polynomial
P (x) = x 4 - 2x 3 + 7x - 6.
(b) If a, b and c are the other
3 zeros, nd the value of
a + b + c and abc.
14. If x = 2 is a double root of
ax 4 - 2x 3 - 8x + 16 = 0, nd the
value of a and the sum of the
other two roots.
15. Two of the roots of
x 3 - px 2 - qx - 4 = 0 are 3 and 5.
(a) Find the other root.
(b) Find p and q.
16. The product of two of the roots
of x 4 + 2x 3 - 18x - 5 = 0 is - 5.
Find the product of the other
two roots.
17. The sum of two of the roots of
x 4 + x 3 + 7x 2 + 14x - 1 = 0 is
4. Find the sum of the other
two roots.
18. Find the roots of
8x 3 - 20x 2 + 6x + 9 = 0, given
that two of the roots are equal.
19. Solve 12x 3 - 4x 3 - 3x + 1 = 0 if
the sum of two of its roots is 0.
20. Solve
6x 4 + 5x 3 - 24x 2 - 15x + 18 = 0
if the sum of two of its roots
is zero.
Test Yourself 12
1.
Write p (x) = x 4 + 4x 3 - 14x 2 - 36x + 45

as a product of its factors.
11. Find the x- and y-intercepts of the curve

y = x 3 - 3x 2 - 10x + 24.
2.
If a, b and c are the roots of

x 3 - 3x 2 + x - 9 = 0, find
(a) a + b + c
(b) abc
(c) ab + ac + bc
1 1 1
(d) + +
a b c
12. Divide p (x) = 3x 5 - 7x 3 + 8x 2 - 5

by x - 2, and write p (x) in the form
p (x) = (x - 2) a (x) + b (x) .
3.
4.
5.
A monic polynomial P (x) of degree 3

has zeros - 2,1 and 6. Write down the
polynomial.
(a) Divide
P (x) = x 4 + x 3 - 19x 2 - 49x - 30 by
x 2 - 2x - 15.
(b) Hence, write P (x) as a product of its
factors.
For the polynomial P (x) = x 3 + 2x 2 - 3x,
find
(a) the degree
(b) the coefficient of x
(c) the zeros
(d) the leading term.
6.
Sketch f (x) = (x - 2) (x + 3)2 showing the

intercepts.
7.
If ax 4 + 3x 3 - 48x 2 + 60x = 0 has a

double root at x = 2, find
(a) the value of a
(b) the sum of the other two roots.
8.
Show that x + 7 is not a factor of

x 3 - 7x 2 + 5x - 4.
9.
If the sum of two roots of

x 4 + 2x 3 - 8x 2 - 18x - 9 = 0 is 0, find
the roots of the equation.
10. The polynomial f (x) = ax 2 + bx + c has

zeros 4 and 5, and f (-1) = 60. Evaluate
a, b and c.
13. Solve 2 cos 3 x + cos 2 x - cos x = 0 for

0c# x # 360c.
14. When 8x 3 - 5kx + 9 is divided by x - 2,
the remainder is -1. Evaluate k.
15. Find the zeros of g (x) = - x 2 + 9x - 20.
16. Sketch P (x) = 2x (x - 3) (x + 5), showing
intercepts.
17. Find the value of k if the remainder is
- 4 when x 3 + 2x 2 - 3x + k is divided by
x - 2.
18. The sum of 2 roots of
x 4 - 7x 3 + 5x 2 - x + 3 = 0 is 3. Find the
sum of the other 2 roots.
19. A polynomial is given by
P (x) = A (x) (x - a) 3 . Show that
P (a) = Pl(a) = 0.
20. Show that x - 5 is a factor of
f (x) = x 3 - 6x 2 + 12x - 35.
21. (a) Show that x - 5 is a factor of
f ] x g = x 3 - 7x 2 - 5x + 75.
(b) Show that f ] 5 g = f l(5) = 0 .
(c) What can you say about the root
at x = 5?
(d) Write f (x) as a product of its factors.
22. The leading term of a polynomial is 3x 3
and there is a double root at x = 3.
Draw an example of a graph of the
polynomial.
713
714
23. A polynomial P(x) has a triple root at

x = - 6.
(a) Write an expression for P(x).
(b) If P(x) has leading coefcient 3 and
degree 4, draw a sketch showing this
information.
24. Draw an example of a polynomial with

leading term 3x 5 .
25. If P ] x g = ax 3 + bx 2 + cx + d has a
remainder of 8 when divided by
x - 1, P ] 2 g = 17, P ] -1 g = - 4 and
P ] 0 g = 5, evaluate a, b, c and d.
Challenge Exercise 12
1.
Write P (x) = x 5 + 2x 4 + x 3 - x 2 - 2x - 1
as a product of its factors.
8.
Find the value of a if (x + 1) (x - 2) is a

factor of 2x 3 - x 2 + ax - 2.
2.
A polynomial P (x) = (x - b) 7 Q (x) .

(a) Show that P (b) = P l(b) = 0.
(b) Hence nd a and b, if (x - 1) 7 is a
factor of
P (x) = x 7 + 3x 6 + ax 5 + x 4 + 3x 3 + bx 2 x +1.
9.
Prove that if x - a is a factor of

polynomial P (x), then P (a) = 0.
Solve
tan 4 i - tan 3 i - 3 tan 2 i + 3 tan i = 0
for 0c# i # 360c.
11. Write down an example of a polynomial

with the graph below.
3.
4.
(a) Find the equation of the tangent to

the curve y = x 3 at the point where x = 1.
(b) Find the point where this tangent
cuts the curve again.
5.
(a) Find the remainder when

p (x) = 2x 4 - 7x 3 + ax 2 + 3x - 9 is
divided by 2x - 1.
(b) If the remainder, when p (x) is
divided by x + 2, is 17, nd the value
of a.
6.
If a, b and c are roots of the cubic

equation 2x 3 + 8x 2 - x + 6 = 0, nd
(a) abc
(b) a2 + b2 + c2
7.
Solve 4 sin 3 i - 3 sin i - 1 = 0 for

0c# i # 360c.
10. Find the points of intersection between

the polynomial P ] x g = x 3 + 5x 2 + 4x - 1
and the line 3x + y + 4 = 0.
-1
12. Sketch an example of a polynomial with

a double root at x = a 1 and a double root
at x = a 2, if the polynomial is monic and
has even degree _ a 2 2 a 1 i .

ch12 PDF

Uploaded by

Copyright:

Available Formats

ch12 PDF

Uploaded by

Document Information

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

ch12 PDF

Uploaded by

Copyright:

Available Formats

12

Monic polynomial: A polynomial where the leading

Degree: The value of the highest power of x in a

Polynomial: A sum or difference of terms involving

Dividend: The number, algebraic expression or polynomial

Quotient: The result when two numbers, algebraic

DID YOU KNOW?

P(x) is a continuous and differentiable function.

The degree of a polynomial

Coefficients can be any real

Maths In Focus Mathematics Extension 1 Preliminary Course

Degree is 6 since x6 is the highest power.

Maths In Focus Mathematics Extension 1 Preliminary Course

Write down the degree of each

Find the zeros of the following

Find the derivative of each

Which of the following are not

For the polynomial

10. Given f (x) = 3x 2 - 2x + 1 and

12. For the polynomial

13. If P ] x g = 3x 4 - 4x 3 - 1, find the

(check this on your calculator)

Maths In Focus Mathematics Extension 1 Preliminary Course

(check this on your calculator)

A polynomial P(x) can be written as P ] x g = A ] x g $ Q ] x g + R ] x g

x - 2 is called the divisor.

Step 2: Multiply 3x 3 by (x - 2) and find the remainder by

Maths In Focus Mathematics Extension 1 Preliminary Course

Continuing this way until we have finished, we will have

The remainder is 67.

(b) This means that (3x 4 - x 3 + 7x 2 - 2x + 3) ' (x - 2)

Check this is true by

This means that

Divide the following polynomials

11. (6x 2 - 3x + 1) ' (3x - 2)

(4x + 2x - 3) ' (2x + 3)

20. (3x 5 - 2x 3 + x - 1) ' (x + 1)

10. (4x 3 - 2x 2 + 6x - 1) ' (2x + 1)

Check this by expanding

Maths In Focus Mathematics Extension 1 Preliminary Course

Remainder and Factor Theorems

If a polynomial P(x) is divided by x - a, then the remainder is P(a)

The remainder when P(x) is divided by x + 3 is P ^ -3 h .

For a polynomial P(x), if P ] a g = 0 then x - a is a factor of the polynomial.

Maths In Focus Mathematics Extension 1 Preliminary Course

Further properties of a polynomial

If polynomial P(x) has k distinct zeros a 1, a 2, a 3, ... a k, then

If polynomial P(x) has degree n and n distinct zeros a 1, a 2, a 3, ... a n,

A polynomial of degree n cannot have more than n distinct real zeros.

A polynomial of degree n with more than n distinct real zeros is the

Maths In Focus Mathematics Extension 1 Preliminary Course

Suppose P(x) has another distinct zero a n + 1

You learned a special case of this result

than n distinct x values.

Substitute (2) into (3):

If x - a is a factor of polynomial P(x), then a is a factor of the constant

We already use this when

Maths In Focus Mathematics Extension 1 Preliminary Course

f (2) = 2 3 + 3 (2) 2 - 4 (2) - 12