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TERMINOLOGY

12 Polynomials 1

Coeffi cient: A constant multiplied by a pronumeral in an algebraic term e.g. in ax3 the a is the coeffi cient

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

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 ( ) 0P a = then x a- is a factor of P(x)

Leading term: The term with the highest power of x. e.g. 5 2 3x x3 2

- + has a leading term of 5x3

Long division: A division of one polynomial into another polynomial using a method similar to long division of numbers

Monic polynomial: A polynomial where the leading coeffi cient is 1

Polynomial: A sum or difference of terms involving integral powers of a variable, usually x. A function of the form P x a a x a x a x( )

0 1 2

2

n

nf= + + + + where a0, a1, ... are real numbers and n is a positive integer or zero

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 ( ) 0P x = . Graphically it is where the polynomial crosses the x-axis.

Zeros: The zeros of a polynomial are the roots of the polynomial equation ( ) 0P x = . They are the values that make P(x) zero.

ch12.indd 662 7/25/09 11:41:17 AM

663Chapter 12 Polynomials 1

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-’?

P x p p x p x p x p xnn

nn

0 1 22

11f= + + + + +

-

-] g where n is a positive integer or zero.

P ( x ) is a continuous and differentiable function. Although the defi nition has the term p n x n last, we generally write

polynomials from the highest order down to the lowest. e.g. .f x x x5 42= - +] g We can describe various aspects of polynomial as follows:

p x p x p x p x p x pnn

nn

nn

11

22

22

1 0f+ + + + + +-

-

-

- is called a polynomial expression P x p x p x p x p x p x pn

nn

nn

n1

12

22

21 0f= + + + + + +

-

-

-

-] g has degree n where 0pn! , , ,p p p p n n n1 2 0f

- - are called coeffi cients

p n x n is called the leading term and p n is the leading coeffi cient p 0 is called the constant term If 1,p P xn = ] g is called a monic polynomial If p p p p 0n0 1 2 f= = = = = then P ( x ) is the zero polynomial

The degree of a polynomial is the highest power of x

with non-zero coeffi cient.

Coeffi cients can be any real number but we generally

use integers in this course for convenience.

Defi nition of a Polynomial

A polynomial is a function defi ned for all real x involving positive powers of x in the form:

ch12.indd 663 7/10/09 11:58:35 PM

664 Maths In Focus Mathematics Extension 1 Preliminary Course

EXAMPLES

1. Which of the following are polynomial expressions? (a) x x4 3 2- + (b) x x x3 5 14 2- + - (c) x x x32 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 x x x x x2 3 7 36 4 3 2= - + + - -] g Find the degree. (a) Is the polynomial monic? (b) State the leading term. (c) What is the constant term? (d) Find the coeffi cient of (e) x 4.

Solution

Degree is 6 since (a) x 6 is the highest power. Yes, the polynomial is monic since the coeffi cient of (b) x 6 is 1. The leading term is (c) x 6 . The constant term is (d) .3- The coeffi cient of (e) x 4 is .2-

Polynomial equation

0P x =] g 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 x x52= -] g

Solution

To fi nd the zeros of the polynomial, we solve .P x 0=] g

x xx x

5 05 0

2 - =

- =] g

ch12.indd 664 7/11/09 8:53:12 PM


,x x

x

0 5 0

5

= - =

=

So the zeros are 0, 5.

2. Find the roots of the polynomial equation .x x x2 3 03 2- - =

Solution

x x x

x x xx x x

2 3 0

2 3 03 1 0

3 2

2

- - =

- - =

- + =

^

] ]

h

g g

, ,x x x0 3 0 1 0= - = + =

,x x3 1= = -

The roots are , , .x 0 3 1= -

3. Show that the polynomial p x x x 42= - +] g has no real zeros.

Solution

We look at the polynomial equation .p x 0=] g

x x 4 02 - + =

The discriminant will show whether the polynomial has real zeros.

b ac4 1 4 1 41 16

15

0

2 2

1

- = - -

= -

= -

] ] ]g g g

So the polynomial has no real zeros.

4. For the polynomial P x ax x x x3 7 15 4 3= - + - +] g Evaluate (a) a if the polynomial is monic. Find the degree of the derivative (b) ( ) .xPl

Solution

For a monic polynomial, (a) 1a = (b) ax x x5 12 3 7= - + -P x 4 3 2l^ h

( )xPl has degree 4 (highest power).

ch12.indd 665 7/10/09 11:58:38 PM


1. Write down the degree of each polynomial.

(a) 5 3 2 3 1x x x x7 5 3- + - + (b) 3 2x x x x2 3 4+ + - + (c) 3 5x + (d) 5 4x x11 8- + (e) 2 5 3x x x2 3- - + 3 (f) (g) 2x x4 -

2. For the polynomial ( ) ,P x x x x7 13 2= - + - fi nd

(a) ( )P 2 (b) ( )P 1- (c) ( )P 0

3. Given ( )P x x 5= + and ( ) ,Q x x2 1= - fi nd

(a) ( )P 11- (b) ( )Q 3 (c) ( ) ( )P Q2 2+ - the degree of (d) ( ) ( )P x Q x+ the degree of (e) ( ) ( )P x Q x$

4. For the polynomial ( ) 3 5 4,P x x x x5 4= - - + fi nd

the degree of (a) ( )P x the constant term (b) the coeffi cient of (c) x4 the coeffi cient of (d) x2

5. Find the zeros of the following polynomials.

(a) ( ) 9P x x2= - (b) ( ) 5p x x= + (c) ( ) 2f x x x2= + - (d) ( ) 8 16P x x x2= - +

(e) ( ) 2 5g x x x x3 2= - +

6. Find the derivative of each polynomial ( )P x and state the degree of ( )xPl

(a) ( ) 3 2 4 5P x x x x x4 3 2= - - + - (b) ( ) 5 3P x x2= + (c) ( ) 9 7 8P x x x x12 5= - + (d) ( ) 3 7 3P x x x x x7 3 2= - + - - (e) ( ) 8 5P x x= +

7. Which of the following are not polynomials?

(a) 5 3 1x x x x4 2- + +

(b) 3x2 x+ (c) 3 7x x2 + - (d) 3 5x + 0 (e) (f) x x3

21 12 - +

(g) x x4 7 53 2+ +-

8. For the polynomial ( ) ( 1) ( 7) 5,P x a x b x c3 2= + + - + + fi nd values for a , b or c if

(a) ( )P x is monic the coeffi cient of (b) x2 is 3 the constant term is (c) 1- (d) ( )P x has degree 2 the leading term has a (e)

coeffi cient of 5

9. Given ( ) 2 5,P x x= + ( ) 2Q x x x2= - - and ( ) 9 ,R x x x3= + fi nd

any zeros of (a) ( )P x the roots of (b) ( ) 0Q x = the degree of (c) ( ) ( )P x R x+ the degree of (d) ( ) ( )P x Q x$ the leading term of (e) ( ) ( )Q x R x$

10. Given ( ) 3 2 1f x x x2= - + and ( ) ,g x x3 3= -

show (a) ( )f x has no zeros fi nd the leading term of (b)

( ) ( )f x g x$ fi nd the constant term of (c)

( ) ( )f x g x+ fi nd the coeffi cient of (d) x in

( ) ( )f x g x$ fi nd the roots of (e)

( ) ( ) 0f x g x+ =

11. State how many real roots there are for each polynomial equation .P x 0=] g

(a) P x x 92= -] g (b) 4P x x2= +] g (c) P x x x3 72= - -] g

12.1 Exercises

ch12.indd 666 7/10/09 11:58:38 PM


(d) 2 3P x x x2= + +] g (e) P x x x3 5 22= - -] g (f) P x x x x x1 4 6= - + +] ] ] ]g g g g (g) P x x x x1 2 5= + - -] ] ] ]g g g g

12. For the polynomial ,P x x x x2 3 36 173 2 -= + +] g fi nd the roots of the derivative polynomial equation ( ) .x 0=Pl

13. If ,P x x x3 4 14 3= - -] g fi nd the zeros of ( ) .xPl

14. Show that ( )x 0=Pl has no real roots if .P x x x x93 2= - +] g

15. Show that ( )Q x 0=l has equal roots given .Q x x x x3 3 53 2= - + +] g

Class Investigation

Here are two examples of long division.

Divide 5715 by 48. 1.

48 5715

48

91

48

435

432

3

119 3rg

This means 48

5715 119483

= +

11948 48 4848

5715483

# # #= +

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.

Divide 4871 by 35. 2.

35 4871

35

137

105

321

315

6

139 6rg

CONTINUED

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.

ch12.indd 667 7/10/09 11:58:39 PM


This means 35

4871 139356

= +

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. 2048 15' 2. 5876 17' 3. 3546 21' 4. 2992 33' 5. 8914 19'

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

( )( )

( )( )( )

A xP x

Q xA xR x

= + where Q ( x ) is the quotient and R ( x ) is the remainder.

( )( )

( )( )( )

( ) ( ) ( )A xP x

Q xA xR x

A x A x A x# # #= +

P x A x Q x R x$= +] ] ] ]g g g 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 ).

( )( )

( )( )

( )

So

This gives .

A xR x

Q xA x

R x

R x A x Q x R x

R x

P x A x Q x R x

1

1

1 1

2

2

$

$

= +

= +

=

= +

] ] ] ]

]

] ] ] ]

g g g g

g

g g g g

ch12.indd 668 7/10/09 11:58:39 PM


EXAMPLES

1. (a) Divide ( ) 3 7 2 3P x x x x x4 3 2= - + - + by 2.x - (b) Hence write ( )P x in the form ( ) ( ) ( ) ( ) .P x A x Q x R x= + (c) Show that (2)P is equal to the remainder.

Solution

(a) Step 1: Divide the leading term by x . x x x3 3i.e. 4 3

' =

2 3 7 2 3x x x x x

3

4 3 2- - + - +

3xg

Step 2: Multiply 3x3 by ( 2)x - and fi nd the remainder by subtraction. ( )x x x x3 2 3 6i.e. 3 4 3- = -

x x x x x

xx x

2 3 7 2 3

53 6

3

4 3 2

3

4 3

- - + - +

-

x3g

Step 3: Bring down the 7x2 and divide 5x3 by x .

x x5 7+

x x x x xx x

2 3 7 2 33 6

3 2

4 3 2

4 3

3 2

- - + - +

-

x x3 5+

g

Step 4: Multiply 5x2 by ( 2)x - and fi nd the remainder by subtraction. x x x x5 2 5 10i.e. 2 3 2- = -] g

x x x x xx x

x x

x x

x

2 3 7 2 33 6

5 7

5 10

17

3 2

4 3 2

4 3

3 2

3 2

2

- - + - +

-

+

-

x x3 5+

g

x 2- is called the divisor.

CONTINUED

ch12.indd 669 7/13/09 12:19:19 PM


Continuing this way until we have fi nished, we will have

x

x xx

x

x x x x

x x

x x

x x

x x

2 3 7 2 3

3 6

5 7

5 10

17 2

17 3432 3

32 6467

3 2

4 3 2

4 3

3 2

3 2

2

2

- - + - +

-

+

-

-

-

+

-

x3 + x5 + 32+x17g

(b) This means that ( ) ( )x x x x x3 7 2 3 24 3 2'- + - + -

( ),x x x3 5 17 323 2= + + + remainder 67

or ( )( )

xx x x x x x x

xx x x x x x x x

23 7 2 3 3 5 17 32

267

3 7 2 3 2 3 5 17 32 67

i.e.4 3 2

3 2

4 3 2 3 2

-

- + - += + + + +

-

- + - + = - + + + +

( ) ( ) ( ) ( )P x A x Q x R xi.e. = + where ( )A x is the divisor, ( )Q x is the quotient and ( )R x is the remainder.

( ) ( ) ( ) ( )

( ) .

P

P

2 3 2 2 7 2 2 2 348 8 28 4 3

67

2

(c)

is equal to the remainder

4 3 2

`

= - + - +

= - + - +

=

2. Divide 3 4x x x3 2- + + by .x x2 -

Solution

x x x x x

x x

x x

x xx

3 4

2

2 24

2 3 2

3 2

2

2

- - + +

-

- +

- +

- +

x 2-

g

This means that

( ) ( ) ( ),

2

3 4 ( 2)( ) ( 4)

x x x x x x x

x xx x x x

x xx

x x x x x x x

3 4 2 4

3 4 4

remainder

i.e.

or

3 2 2

2

3 2

2

3 2 2

'- + + - = - - +

-

- + += - +

-

- +

- + + = - - + - +

The remainder is 67.

The quotient is 3x 5x 17x 32.

3 2+ + +

Check this is true by expanding and simplifying.

ch12.indd 670 7/10/09 11:58:41 PM


3. Divide 5 6 15x x x x5 3 2+ + - + by 3.x2 +

Solution

x x x x x

x x

x x x

x x

x

x

3 5 6 15

3

2 5 6

2 6

5 15

5 150

3

2 5 3 2

5 3

3 2

3

2

2

+

+

+ + + - +

+

- + -

- -

x x2 5- +

g

This means that

( ) ( ) ( )

2 5

5 6 15 ( 2 5)( 3)

x x x x x x x

xx x x x x x

x x x x x x x

5 6 15 3 2 5

35 6 15i.e.

or

5 3 2 2 3

2

5 3 23

5 3 2 3 2

'+ + - + + = - +

+

+ + - += - +

+ + - + = - + +

0,R (x) = so there is no remainder.

Check this by expanding and simplifying.

Divide the following polynomials and put them in the form ( ) ( ) ( ) ( ) .P x A x Q x R x= +

1. (3 2 5) ( 4)x x x2'+ + +

2. ( 7 4) ( 1)x x x2'- + -

3. ( ) ( )x x x x2 1 33 2'+ + - -

4. (4 2 3) (2 3)x x x2'+ - +

5. ( 5 2) ( 3 )x x x x x3 2 2'- + + +

6. ( 3) ( 2)x x x x3 2'+ - - -

7. ( ) ( )x x x x x5 2 3 13 2 2'- + + +

8. ( 2 3) ( 4)x x x x x4 3 2'- - + - +

9. (2 5 2 2 5)

( 2 )

x x x x

x x

4 3 2

2

'- + + -

-

10. ( ) ( )x x x x4 2 6 1 2 13 2'- + - +

11. ( ) ( )x x x6 3 1 3 22'- + -

12. ( 2 2) ( )x x x x x4 3 2 2'- - - -

13. ( )( )

x x x x xx3 2 3 1

2

5 4 3 2'- - + - -

+

14. ( 5 2) ( 1)x x x2'+ - +

15. ( 2 5 4) ( 3)x x x x4 2'- + + -

16. (2 5) ( 2 )x x x x4 3 2'- + -

17. ( ) ( )x x x x3 3 1 53 2 2'- + - +

18. (2 4 8) ( 3 2)x x x x x3 2 2'+ - + + +

19. ( 2 4 2 5)

( 2 1)

x x x x

x x

4 3 2

2

'- + + +

+ -

20. ( ) ( )x x x x3 2 1 15 3'- + - +

12.2 Exercises

ch12.indd 671 7/10/09 11:58:41 PM


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

Proof

P x A x Q x R x$= +] ] ] ]g g g g where A x x a= -] g P x x a Q x R x= - +] ] ] ]g g g g The degree of A ( x ) is 1, so the degree of R ( x ) must be 0 .

So where is a constant

Substituting :

So is the remainder.

kR x k

P x x a Q x kx a

P a a a Q a k

Q x kk

P a

0

`

$

=

= - +

=

= - +

= +

=

]

] ] ]

] ] ]

]

]

g

g g g

g g g

g

g

The degree of R(x) is less than the degree of A(x).

EXAMPLES

1. Find the remainder when x x x3 2 5 14 2- + + 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 3 2 2 2 5 2 1

51

4 2= - + +

=

] ] ] ]g g g g

So the remainder is 51.

2. Evaluate m if the remainder is 4 when dividing 2 5x mx4 + + by 3x + .

Solution

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

4P

mm

3

2 3 3 5 4162 3 5 4

oS4

- =

- + - + =

- + =

]

] ]

g

g g

x 3 x ( 3) .=+ - -

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

ch12.indd 672 7/10/09 11:58:41 PM


Factor theorem

The factor theorem is a direct result of the remainder theorem.

.

m

m

m

m

167 3 4

167 3 4

163 3

5431

- =

= +

=

=

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

Proof

P x A x Q x R x$= +] ] ] ]g g g g where A x x a= -] g P x x a Q x R x= - +] ] ] ]g g g g The remainder when P ( x ) is divided by x a- is P ( a ) . So P x x a Q x P a= - +] ] ] ]g g g g But if 0:P a =] g

P x x a Q x

x a Q x

0= - +

= -

] ] ]

] ]

g g g

g 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 0P a =] g

Proof

If x a- is a factor of P ( x ), then we can write : P x x a Q x= -] ] ]g g g This means that when P ( x ) is divided by ,x a- the quotient is Q ( x ) and there is no remainder. So 0P a =] g

ch12.indd 673 7/10/09 11:58:42 PM


Further properties of a polynomial

Some properties of polynomials come from the remainder and factor theorems.

EXAMPLE

(a) Show that x 1- is a factor of P x x x x7 8 23 2= - + -] g . (b) Divide P ( x ) by x 1- and write P ( x ) in the form P x x Q x1= -] ] ]g g g .

Solution

(a) The remainder when dividing the polynomial by x 1- is P (1)

P 1 1 7 1 8 1 2

0

3 2= - +

=

-] ] ]g g g

So x 1- is a factor of P ( x ) .(b)

x 1- x x x

x x

x x

x xx

x

0

7 8 2

6 8

6 62 2

2 2

3 2

2

3 2

2

2

- + -

-

- +

- +

-

-

x - 6 2x +

g

So P x x x x

x x x

7 8 2

1 6 2

3 2

2

= - + -

= - - +

]

] ^

g

g h

Notice that x 6x 22 - + won’t factorise.

If polynomial P ( x ) has k distinct zeros , , , ...a a a ak1 2 3 , then ( ) ( ) ( ) ... ( )x a x a x a x ak1 2 3- - - - is a factor of P ( x )

Proof

If a 1 is a zero of P ( x ) then ( )x a1- is a factor of P ( x ) . If a 2 is a zero of P ( x ) then ( )x a2- is a factor of P ( x ) . If a 3 is a zero of P ( x ) then ( )x a3- is a factor of P ( x ) . Similarly, if ak is a zero of P ( x ) then ( )x ak- is a factor of P ( x ) .

So( )( ) ( ) ... ( )

( ) ( ) ( ) ... ( ) is a factor ofP x x a x a x a x a g xx a x a x a x a P x

k

k

1 2 3

1 2 3

` = - - - -

- - - - .] ]

]

g g

g

If polynomial P ( x ) has degree n and n distinct zeros , , , ...a a a an1 2 3 , then ( ) ( ) ( ) ... ( )P x p x a x a x a x an n1 2 3= - - - -] g

ch12.indd 674 7/10/09 11:58:42 PM


Proof

Since , , , ...a a a an1 2 3 are zeros of P ( x ), ( ) ( ) ( ) ... ( )x a x a x a x an1 2 3- - - - is a factor of the polynomial. So ( ) ( ) ( ) ... ( )P x x a x a x a x an1 2 3= - - - -] g Q ( x ) But ( ) ( ) ( ) ... ( )x a x a x a x an1 2 3- - - - has degree n and P ( x ) has degree n so Q ( x ) must be a constant . ( ) ( ) ( ) ... ( )P x p x a x a x a x an n1 2 3` = - - - -] g

EXAMPLE

If a polynomial has degree 2, show that it cannot have 3 zeros .

Solution

Let P x p x p x p22

1 0= + +] g where 0p2! Assume P ( x ) has 3 zeros, a 1 , a 2 and a 3 Then x a x a x a1 2 3- - -_ _ _i i i is a factor of the polynomial. ( ) ( )( )( ) ( )P x x a x a x a Q x1 2 3` = - - - 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 cannot have more than n distinct real zeros.

A polynomial of degree n with more than n distinct real zeros is the zero polynomial ( ...P x p p p0 0n0 1 2= = = = = =p] g )

Proof

P ( x ) has degree n So P x p x p x p x p x p x pn

nn

nn

n1

12

22

21 0f= + + + + + +

-

-

-

-] g where 0pn! Suppose P ( x ) has more than n distinct zeros, say 1n + Then ( ) ( ) ( ) ... ( )x a x a x a x an1 2 3 1- - - -

+ is a factor of P ( x ).

So ( ) ( ) ( ) ... ( )P x x a x a x a x a Q xn1 2 3 1= - - - -+

] ]g g . But this gives P ( x ) at least degree 1,n + 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.

Proof

Let P ( x ) be a polynomial of degree n with zeros , , , ...a a a an1 2 3 Then ( ) ( ) ( ) ... ( )P x x a x a x a x a kn1 2 3= - - - -] g

ch12.indd 675 7/10/09 11:58:42 PM


Suppose P ( x ) has another distinct zero a 1n+ Then 0P a 1n =

+_ i ( ) ( ) ( ) ... ( )a a a a a a a a k 0n n n n n1 1 1 2 1 3 1` - - - - =

+ + + +

But , , , ...a a a a an n1 1 2 3!+

since all zeros are distinct. So 0k = 0P x` =] g

If two polynomials of degree n are equal for more than n distinct values of x , then the coeffi cients of like powers of x are equal. That is, if ... ...a a x a x a x b b x b x b xn

nn

n0 1 2

20 1 2

2/+ + + + + + + + then , , , ...a b a b a b a bn n0 0 1 1 2 2= = = =

Proof

Let ...A x a a x a x a xnn

0 1 22= + + + +] g

and ...B x b b x b x b xnn

0 1 22= + + + +] g

where A x B x=] ]g g for more than n distinct x values . Let P x A x B x= -] ] ]g g g Then ( ) ( ) ( ) ... ( )P x a b a b x a b x a b xn n

n0 0 1 1 2 2

2= - + - + - + + -] g and P ( x ) has degree n . If A x B x=] ]g g for more than n distinct x values, then A x B x 0=-] ]g g for more than n distinct x values . This means 0P x =] g for more than n distinct x values . This means that P ( x ) has more than n zeros. P ( x ) is the zero polynomial 0P x =] g ( ) ( ) ( ) ... ( )a b a b x a b x a b x 0n n

n0 0 1 1 2 2

2- + - + - + + - = So , , , ...,a b a b a b a b0 0 0 0n n0 0 1 1 2 2- = - = - = - = , , , ...,a b a b a b a bn n0 0 1 1 2 2` = = = =

You learned a special case of this result in Chapter 10 under quadratic identities. This is a more general result for all polynomials.

EXAMPLE

Write x x2 53 2 +- in the form 3 3ax b x c x d3 2+ + + + +] ]g g .

Solution

( ) ax b x c x d ax b x x c x dax bx bx b cx c d

ax bx b c x b c d

3 3 6 9 36 9 3

6 9 3

3 2 3 2

3 2

3 2

+ + + + + = + + + + + +

= + + + + + +

= + + + + + +

] ^ ]

]

g h g

g

For x x ax b x c x d2 5 3 33 2 3 2/+ + + + + +- ] ]g g

ch12.indd 676 7/13/09 12:20:02 PM


Proof Let P x p x p x p x p x p x pn

nn

nn

n1

12

22

21 0f= + + + + + +

-

-

-

-] g where p 0n!

If x a- is a factor of P ( x ) we can write P x x a Q x= -] ] ]g g g where Q ( x ) has degree .n 1-

P x x a q x q x q x q x q q

xq x xq x xq x xq

aq x aq x aq x aq x aq

q x q x q x q x aq x aq x

aq x aq x aq

q x q a x q a x q a x aq

0wherenn

nn

n

nn

nn

nn

nn

nn

nn

nn

nn

nn

nn

11

22

22

1 0 1

11

22

1 0

11

22

22

1 0

1 21

12

0 11

22

22

1 0

1 21

12

0 0

g

g

g

g g

g

!= - + + + + +

= + + + +

- - - - - -

= + + + + - - -

- - -

= + - + + - + - -

-

-

-

-

-

-

-

-

-

-

-

-

-

- -

-

-

-

-

-

- -

-

] ] _

_ _ _

g g i

i i i

p aq0 0` = - So a is a factor of p 0 .

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

We already use this when factorising a trinomial. This is a more general

result for all polynomials.

a

b

b c

b c d

1 1

2 2

6 0 3

9 3 5 4

=

= -

+ =

+ + =

]

]

]

]

g

g

g

g

Substitute (2) into (3):

cc

c

6 2 012 0

12

- + =

- + =

=

] g

Substitute 2b = - and 12c = into (4):

.

dd

d

d

x x x x x

9 2 3 12 518 36 5

18 5

13

2 5 2 3 12 3 1323 3 2` /

- + + =

- + + =

+ =

= -

- + - + + + -

] ]

] ]

g g

g g

ch12.indd 677 7/10/09 11:58:43 PM


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 x x3 4 123 2= + - -

Solution

Try factors of ( , , , , , ) .12 1 2 3 4 6 12i.e. ! ! ! ! ! !-

0!

( ) ( ) ( )f 1 1 3 1 4 1 1212

e.g. 3 2= + - -

= -

x 1` - is not a factor of ( )f x

( ) ( ) ( )f 2 2 3 2 4 2 12

0

3 2= + - -

=

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 2x - to fi nd other factors:

x x5x x x x

x x

x x

x xx

x

0

2 3 4 122

5 4

5 106 12

6 12

2

3 2

3 2

2

2

- + - -

-

-

-

-

-

+ + 6g

x 2- is called a linear factor as it has degree 1.

EXAMPLE

Factorise x x2 152 + - .

Solution

Factors of 15- are , , ,3 5 3 5 1 15 1 15# # # #- - - - . We choose 3 5#- since ,x x x3 5 2- + = the middle term.

So x x x x2 15 3 52 + =- - +] ]g g .

ch12.indd 678 7/25/09 11:41:21 AM


( ) ( ) ( )

( ) ( ) ( )f x x x x

x x x2 5 62 2 3

2` = - + +

= - + +

2. Find all factors of ( ) 3 5 15.P x x x x3 2= + + +

Solution

Try factors of 15 ( , , , ) .1 3 5 15i.e.! ! ! !

( 3) ( 3) 3( 3) 5( 3) 15P

0e.g. 3 2- = - + - + - +

=

3x` + is a factor of ( )f x

We divide ( )P x by 3x + to fi nd other factors:

x

0

3+

x

x x xx x

x

x

5

3 5 153

0 5 15

5 15

2

3 2

3 2

+

+ + +

+

+ +

+

g

( ) ( ) ( )P x x x3 52` = + +

x 5+2 will not factorise for any real x.

1. Use the remainder theorem to fi nd the remainder in each question.

(a) ( ) ( )x x x x2 5 43 2'- + + -

(b) ( ) ( )x x x5 3 22'+ + +

(c) ( ) ( )x x x2 4 1 33'- - +

(d) ( ) ( )x x x x3 2 4 55 2'+ - + -

(e) ( ) ( )x x x x5 2 2 9 13 2'+ + - -

(f) ( ) ( )x x x x x3 1 24 3 2'- + - - +

(g) ( ) ( )x x x2 7 2 72'+ - +

(h) ( ) ( )x x x5 1 37 3'+ - -

(i) ( ) ( )x x x x2 3 4 56 2'- + + +

(j) ( ) ( )x x x x x3 7 14 3 2'- - - - +

2. Find the value of k if the remainder is 3 when (a)

x x k5 102 - + is divided by .x 1- the remainder is (b) 4- when

( )x k x kx1 5 43 2- - + + is divided by 2x + .

the remainder is 0 when (c) x x k2 7 15 2+ + + is divided by 6x + .

(d) x kx x x2 34 3 2- + + is divisible by x 3- .

the remainder is 25 when (e) x x2 3 54 2- + is divided by x k- .

3. (a) Find the remainder when ( )f x x x x4 63 2= - + + is divided by .x 2-

Is (b) x 2- a factor of ( )?f x Divide (c) x x x4 63 2- + +

by 2.x - Factorise (d) ( )f x fully and write

( )f x as a product of its factors.

4. (a) Show that 3x + is a factor of ( ) .P x x x x x3 9 274 3 2= + - -

Divide (b) ( )P x by 3x + and write ( )P x as a product of its factors.

12.3 Exercises

ch12.indd 679 7/10/09 11:58:44 PM


5. The remainder is 5 when ( ) 4 4P x ax bx x3 2= - + - is divided by 3x - 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 x3 12= - + and ( )g x x x3 23 2= - + are divided by x 1+ they leave the same remainder. Find the value of a .

7. (a) Show that 3x - is not a factor of ( ) .P x x x x x2 7 3 55 4 2= - + - +

Find a value of (b) k such that 3x - is a factor of ( ) .Q x x x k2 53= - +

8. The polynomial ( ) 2P x x ax bx3 2= + + + has factors x 1+ and .x 2- Find the values of a and b .

9. (a) The remainder, when ( ) 15 9 2f x ax bx x x4 3 2= + + + + is divided by 2,x - is 216, and x 1+ is a factor of ( ) .f x Find a and b .

Divide (b) ( )f x by x 1+ and write the polynomial in the form ( ) ( ) ( ) .f x x g x1= +

Show that (c) x 1+ is a factor of ( ) .g x

Write (d) ( )f x as a product of its factors.

10. Write each polynomial as a product of its factors .

(a) 2 8x x2 - -

(b) 2x x x3 2+ -

(c) 10 8x x x3 2+ - +

(d) 4 11 30x x x3 2+ - -

(e) 11 31 21x x x3 2- + -

(f) 12 17 90x x x3 2- + +

(g) 7 16 12x x x3 2- + -

(h) 6 9 4x x x x4 3 2+ + +

(i) 3 4x x3 2+ -

(j) 7 6x x3 - -

11. (a) Write ( ) 7 6P x x x3= - + as a product of its factors.

What are the zeros of (b) ( )?P x Is (c) ( 2)( 3)x x- + a factor

of ( )?P x

12. If ( ) 10 23 34 120f x x x x x4 3 2+ += - -

has zeros 5- and 2 show (a) ( 5)( 2)x x+ - is a factor

of ( )f x write (b) ( )f x as a product of its

linear factors.

13. If ( ) 3 13 51 36P x x x x x4 3 2+= - - - has zeros 3- and 4, write ( )P x as a product of its linear factors.

14. (a) Show that ( ) 3 34 120P x x x x3 2= - - + has zeros 6- and 5.

Write (b) ( )P x as a product of its linear factors.

15. (a) Write the polynomial P u u u u4 5 23 2= - + -] g as a product of its factors.

Hence or otherwise, solve (b) .x x x1 4 1 5 1 2 03 2- - - + - - =] ] ]g g g

16. (a) Write the polynomial f p p p p2 5 63 2= - - +^ h as a product of its factors .

(b) Hence or otherwise, solve .x x x2 1 2 2 1 5 2 1 6 03 2+ - + - + + =] ] ]g g g

17. (a) Write P k k k2 3 13 2= + -] g as a product of its factors.

(b) Hence or otherwise, solve sin sinx x2 3 1 03 2+ - = for .c cx0 360# #

18. (a) Write 13 39 27f u u u u3 2= - + -] g as a product of its factors.

(b) Hence or otherwise, solve . . .3 13 3 39 3 27 0x x x3 2- + - =

19. Solve .x x x4 4 2 4 04 3 2+ - + - + =] ] ]g g g

Linear factors are in the form .x a-

ch12.indd 680 7/10/09 11:58:44 PM


20. Solve cos cos cos2 03 2i i i- - = for x0 360c c# # .

21. Evaluate a , b , c and d if (a) x x x3 2 13 2 /+ - +

ax b x cx d13 2+ - + +] g

(b) x x x43 2 /- +

ax b x c x d2 23 2+ + + + +] ]g g

(c) x x2 73 /- +

ax b x c x d1 1 23 2+ + + + + +] ]g g

(d) x x x5 33 2 /+ + -

ax b x cx d33 2+ - + +] g

(e) x x4 33 /- +

a x b x1 43 2+ + + +] ]g g

c x d4 1+ + -] g

(f) x x x8 63 2 /+ - -

ax b x cx d2 33 2+ - + + -] g

(g) x x x3 23 2 /- +

a x b x2 53 2- + - +] ]g g

c x d5 2- + -] g

(h) x x x4 23 2 /- + - -

a x bx cx d1 3 2+ + + +] g

(i) x x2 3 13 2 /- + -

ax b x cx d2 13 2+ - + +] g

(j) x x x4 33 2 /- - + +

a x b x2 23 2- + - +] ]g g

c x d2 1- + +] g

22. A monic polynomial of degree 3 has zeros ,3 0- and 4. Find the polynomial.

23. Polynomial P x ax bx cx 83 2= - + -] g has zeros 2 and 1- and .P 3 28=] g Evaluate a , b and c.

24. A polynomial with leading term 2 x 4 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 6P x x x x3 2= + -] g as a product of its factors. (b) Sketch the graph of the polynomial.

Solution

(a) P x x x x

x x xx x x

6

63 2

3 2

2

= + -

= + -

= + -

]

^

] ]

g

h

g g

(b) For the graph of P x x x x63 2= + -] g For x -intercepts: 0y =

CONTINUED

ch12.indd 681 7/10/09 11:58:45 PM


, ,

,

x x xx x x

x x x

x x

0 63 2

0 3 0 2 0

3 2

3 2= + -

= + -

= + = - =

= - =

] ]g g

So x -intercepts are 0, 3- and 2.

For y -intercepts: 0x =

P 0 0 0 6 0

0

3 2= + -

=

] ] ]g g g

So y -intercept is 0.

0 2-3 1 3 4-2 -1-4

1

2

3

4

-1

-2

-3

-4

y

x

We look at which parts of the graphs are above and which are below the x -axis between the x -intercepts. , :x x3 4Test say1- = -

P x x x xx x x

P

63 2

4 4 4 3 4 2

4 1 624

0

3 2

1

= + -

= + -

- = - - + - -

= - - -

= -

]

] ]

] ] ]

] ]

g

g g

g g g

g g

So the curve is below the x -axis.

, :x x3 0Test say 11 1- = -

0

P

2 36

1 1 1 3 1 2

1

2

- = - - + - -

= - -

=

] ] ]

] ]

g g g

g g

So the curve is above the x -axis.

ch12.indd 682 7/25/09 11:41:22 AM


, :x x

P

0

0

Test 2 say 1

1 1 1 3 1 2

1 4 14

=

= + -

=

= -

1

1

1

-

] ] ]

] ]

g g g

g g

So the curve is below the x -axis.

, :x xTest 2 say 32 =

P

0

3 3 3 3 3 2

3 6 118

= + -

=

=

2

] ] ]

] ]

g g g

g g


We can sketch the polynomial as shown.

0 2-3 1 3 4-2 -1-4

1

2

3

4

-1

-2

-3

-4

y

x

2. (a) Write the polynomial 5 3P x x x x3 2= - - -] g as a product of its factors. (b) Sketch the graph of the polynomial.

Solution

(a) Factors of 3- are 1! and 3! .

P 1 1 1 5 1 3

0

3 2= - - - - -

=

- -] ] ] ]g g g g

Later on, in a class investigation in this chapter you will learn

how to make the graph more accurate by fi nding the

maximum and minimum points. This is a topic in the HSC Course.

CONTINUED

ch12.indd 683 7/25/09 11:41:22 AM


So 1x + is a factor of the polynomial.

x

x

3x2x x x

x

x x

x xx

x

1 5 3

2 5

2 23 3

3 3

0

2

3 2

3 2

2

2

+ - - -

+

- -

- -

- -

- -

x - -

g

P x x x x

x x x

x x

1 2 31 3 1

1 3

2

2

= + - -

= + - +

= + -

] ] ^

] ] ]

] ]

g g h

g g g

g g

(b) For the graph of 5 3P x x x x3 2= - - -] g

For x-intercepts: y = 0

x 1= -

,

x x x

x x

x x

0 5 3

1 3

1 0 3 0

3 2

2

2

= - - -

= + -

+ = - =

x 1 0+ = x 3=

] ]

]

g g

g

So x -intercepts are 1- and 3 . For y -intercepts: 0x =

P 0 0 0 5 0 3

3

3 2= - - -

= -

] ] ]g g g

So y -intercept is 3- .

We look at which parts of the graphs are above and which are below the x -axis between the x -intercepts. , :x x1 2Test say1- = -

01

5

P x x x x

x x

P

3 3

1 3

2 2 1 2 3

1 5

3 2

2

2

2

= - - +

= + -

- = - + - -

= - -

= -

]

] ]

] ] ]

] ]

g

g g

g g g

g g

So the curve is below the x -axis .

, :x x1 3 0Test say1 1- =

01

P 0 0 1 0 3

1 33

2

2

= + -

= -

= -

] ] ]

] ]

g g g

g g

0 2-3 1 3 4-2 -1-4

1

2

3

4

-1

-2

-3

-4

y

x

ch12.indd 684 7/25/09 5:57:37 PM


So the curve is below the x -axis . , :x x3 4Test say2 =

02

4 1 4 3P 4

5 125

2

2

= +

=

=

-] ] ]

] ]

g g g

g g


We can sketch the polynomial as shown.

0 2-3 1 3 4-2 -1-4

1

2

3

4

-1

-2

-3

-4

y

x

-5

-6

-7

1. Sketch the graph of each polynomial by fi nding its zeros and showing the x - and y -intercepts .

(a) f x x x x1 2 3= + - -] ] ] ]g g g g

(b) 4 2P x x x x= + -] ] ]g g g

(c) 1 3p x x x x= - - -] ] ]g g g

(d) f x x x 2 2= +] ]g g

(e) g x x x x5 2 5= - + +] ] ] ]g g g g

2. (i) Write each polynomial as a product of its factors

Sketch the graph of the (ii) polynomial

(a) 2 8P x x x x3 2= - -] g

(b) 4 5f x x x x3 2= - - +] g

(c) 3 2P x x x x4 3 2= + +] g

(d) 2 15A x x x x3 2= + -] g

(e) 2 3P x x x x4 3 2= - + +] g

12.4 Exercises

ch12.indd 685 7/25/09 11:41:23 AM


3. (a) Find the x -intercepts of the polynomial 3 4P x x x x4 3= + -] g .

Sketch the graph of the (b) polynomial .

4. (a) Show that 2x - is a factor of 3 4 12P x x x x3 2= - - +] g .

Write (b) P ( x ) as a product of its factors .

Sketch the graph of the (c) polynomial .

5. Sketch the graph of each polynomial, showing all x - and y -intercepts .

(a) P x x x x3 10 243 2= + - -] g

(b) P x x x x9 93 2= + - -] g

(c) P x x x x12 19 8 2 3= - + -] g

(d) P x x x13 123= - +] g

(e) P x x x x2 9 183 2= - + + -] g

(f) P x x x x2 4 83 2= + - -] g

(g) P x x x x5 8 43 2= - + -] g

(h) P x x x x5 33 2= + - +] g

(i) ( )f x x x x16 12 2 4= + -

(j) 2 1P x x x4 2= - +] g

Class Investigation

The graphs in the examples above are not very accurate, as we don’t know where they turn around. We can use calculus to help fi nd these points.

You used the axis of symmetry to fi nd the minimum and maximum values of quadratic functions in Chapter 10. You can also use calculus to fi nd the minimum or maximum turning points of functions.

Notice that the graph below has both a maximum and minimum turning point. We can fi nd these by looking at the gradient of the tangents

around the curve, or dx

dy .

y

x

Maximum turning point

Minimum turning point

You will look at the applications of calculus in sketching graphs in the HSC Course.

You looked at the gradient of tangents to a curve in Chapter 8.

ch12.indd 686 7/10/09 11:58:47 PM


Notice that at both these turning points, 0.dx

dy=

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.

So dx

dy02 on the LHS and

dx

dy01 on the RHS .

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.

So dx

dy01 on the LHS and

dx

dy02 on the RHS .

CONTINUED

ch12.indd 687 7/10/09 11:58:48 PM


There is also another type of point that you see in graphs such as f x x3=] g .

This is called a point of infl exion and has dx

dy0= .

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 2 3 12 7P x x x x3 2= + - -] g showing any stationary points.

dx

dyx x6 6 122= + -

For stationary points 0:dx

dy=

,

,

x x

x xx x

x x

x x

6 6 12 0

6 2 06 1 2 0

1 0 2 0

1 2

2

2

+ - =

+ - =

- + =

- = + =

= = -

^

] ]

h

g g

So there are two stationary points when ,x 1 2= - .

x

P

1 1

1 2 1 3 1 12 1 714

When3 2

=

= + - -

= -

]

] ] ] ]

g

g g g g

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.

( ) ( )

( ) ( )

x

dx

dy

x

dx

dy

0

6 0 6 0 12

12

2

6 2 6 2 12

24

When

When

2

2

=

= + -

= -

=

= + -

=

ch12.indd 688 7/25/09 5:57:49 PM


x 0 1 2

dx

dy-12 0 24

Since dx

dy01 on the LHS and

dx

dy02 on the RHS , ,1 14-^ h is a minimum

turning point .

x2 2When = -] g

P 2 2 2 3 2 12 2 713

3 2- = - + - - - -

=

] ] ] ]g g g g

So there is a stationary point at ,2 13-^ h . Check the gradient on the LHS and RHS of this point.

( ) ( )

( ) ( )

x

dx

dy

x

dx

dy

3

6 3 6 3 12

24

1

6 1 6 1 12

12

When

When

2

2

= -

= - + - -

=

= -

= - + - -

= -

x -3 -2 -1

dx

dy24 0 -2

Since dx

dy02 on the LHS and

dx

dy01 on the RHS , ,2 13-^ h is a maximum

turning point .

Now we look for intercepts.

For x -intercepts: 0y = x x x0 2 3 12 73 2= + - - The expression x x x2 3 12 73 2+ - - will not factorise so we cannot fi nd the x -intercepts.

For y -intercept: 0x =

P x 2 0 3 0 12 0 7

7

3 2= + - -

= -

] ] ] ]g g g g

So the y -intercept is 7- .

Factors of 7- are 1! and 7! and none of

these factors will satisfy the polynomial equation.

CONTINUED

ch12.indd 689 7/10/09 11:58:49 PM


We sketch the polynomial using the stationary points and y -intercept.

(1, -14)

(-2, 13)

-7

y

x

Can you sketch the following polynomials using calculus to fi nd their stationary points?

1. P x x x6 32= + -] g

2. P x x x4 12= - + +] g

3. p x x 53= -] g

4. f x x 24= +] g

5. g x x x2 3 13 2= + -] g

6. P x x x x2 21 72 123 2= - + -] g

7. f x x x x2 9 12 43 2= - + - +] g

8. P x x x x3 3 53 2= - + -] g

9. 8 18 7A x x x x4 3 2= + - -] g

10. Q x x x x x3 20 48 48 34 3 2= - + - + -] g

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 xnn.

ch12.indd 690 7/10/09 11:58:49 PM


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 x x x2 3 7 15 2= + - -] g and 2f x x5=] g 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 fi nd the same results?

All positive or negative values of x to an even power will always be

positive.

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 coeffi cient is

positive, then the polynomial will be positive as x becomes large. This means that for any polynomial with a positive leading coeffi cient

and even degree, P x xas" " !3 3] g . On the graph, both ends of the graph will go up as shown by the

examples below.

y

x

y

x

ch12.indd 691 7/10/09 11:58:49 PM


y

x

If the degree of a polynomial P ( x ) is even and the leading coeffi cient is negative, then the polynomial will be negative as x becomes large.

This means that for any polynomial with a negative leading coeffi cient and even degree, P x xas" " !3 3-] g .

On the graph, both ends of the graph will go down as shown by the examples below.

y

x

y

x

ch12.indd 692 7/10/09 11:58:50 PM


y

x

If P ( x ) is an odd degree polynomial with positive leading coeffi cient, 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 xas" "3 3- -] g and .P x xas" "3 3] g 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

x

y

x

ch12.indd 693 7/10/09 11:58:50 PM


y

x

y

x

If P ( x ) is an odd degree polynomial with negative leading coeffi cient, 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 xas" "3 3-] g and .P x xas" "3 3-] g 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

x

ch12.indd 694 7/10/09 11:58:51 PM


y

x

y

x

y

x

ch12.indd 695 7/10/09 11:58:51 PM


If P ( x ) has even degree, the ends of the graph both go the same way.

If P ( x ) has odd degree, the ends of the graph both go different ways .

A polynomial of odd degree always has at least one real zero .

At least one maximum or minimum value of P ( x ) occurs between any two distinct real zeros .

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.

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 0P x =] g .

y

x

Leading coefficient10

yLeading coefficient20

x

y

x

Leading coefficient20 y

x

Leading coefficient10

ch12.indd 696 7/10/09 11:58:52 PM


When the graph passes through one x -intercept, say x 1 , it must turn around again to pass through the other x -intercept x 2 as shown in the examples below. So there must be at least one maximum or minimum value between the zeros.

y

xx1 x2

y

xx1 x2

y

xx1 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).

ch12.indd 697 7/10/09 11:58:52 PM


EXAMPLE

Solve 2 1 0x x2 - + = .

Solution

x xx x

x

2 1 01 1 0

1 0

2

2

- + =

- - =

- =

] ]

]

g g

g

,

,

x x

x x

1 0 1 0

1 1

- = - =

= =

The solution is 1x = but we say that there are two equal roots .

If ,P x x a Q xn= -] ] ]g g g the polynomial has a multiple root at x a= It has n equal roots at x a=

If P ( x ) has two equal roots at x a= then we can write P x x a Q x2= -] ] ]g g g We say that the polynomial has a double root at x a= . If ,P x x a Q x3= -] ] ]g g g the polynomial has a triple root at .x a= There are

three equal roots at .x a=

EXAMPLES

1. Sketch the graph of f x x 2 2+=] ]g g .

Solution

This graph is f x 2x=] g translated 2 units to the left .

Notice that there is a minimum turning point at the root 2x = - .

0 2-3 1-2 -1-4

1

2

3

4

-1

-2

-3

-4

5

y

x

See class investigations on pages 686–690.

ch12.indd 698 7/13/09 12:20:24 PM


2. Sketch the graph of F x x 1 3= -] ]g g .

Solution

This is the graph of F x x3=] g translated 1 unit to the right.

0 2-3 1 3 4-2 -1-4

2

4

6

8

-2

-4

-6

-8

y

x

Notice that there is a point of infl exion at the root x 1= .

EXAMPLE

(a) Examine the polynomial P x x x2 12= + -] ] ]g g 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 x x2 12= + -] ] ]g g g has roots when P x 0=] g .

,

,

x xx x

x x

2 1 02 0 1 0

2 1

2+ - =

+ = - =

= - =

] ]g g

Notice that there is a double root at x 2= - .

Look at the sign of P ( x ) close to 1:x =

When .

. . .

x

P

0 9

0 9 0 9 2 0 9 12

#

=

= + -

= -

= -

+

] ] ]g g g

So the curve is below the x -axis on the LHS .

Generally, a graph cuts the x -axis at a single root but touches the x -axis at a multiple root in a special way.

See class investigations on pages 686–690.

CONTINUED

ch12.indd 699 7/13/09 12:20:41 PM


When .

. . .

x

P

1 1

1 1 1 1 2 1 1 12

#

=

= + -

= + +

= +

] ] ]g g 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

P

2 1

2 1 2 1 2 2 1 12

#

= -

- = - + - -

= + -

= -

] ] ]g g g

So the curve is below the x -axis on the LHS .

When 1.9

. . .

x

P 1 9 1 9 2 1 9 12

#

= -

- = - + - -

= + -

= -

] ] ]g g g

So the curve is below the x -axis on the RHS.

At the single root 1x = , 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.

Expanding (b) P x x x2 12= + -] ] ]g g g gives x 3 as the leading term.

P x x x

x x xx x x x x

2 1

4 4 14 4 4 4

2

2

3 2 2

= + -

= + + -

= - + - + -

] ] ]

^ ]

g g g

h g

So the polynomial has degree 3 since the highest power is x 3 . Also the leading coeffi cient is 1.

Since P ( x ) has odd degree and a positive leading coeffi cient, as x becomes a larger positive number, P x "3] g and as x becomes a larger negative number, P x " 3-] g .

(c)

1-2

y

x

There is no need to expand the brackets fully as we only need the leading term.

ch12.indd 700 7/10/09 11:58:54 PM


If P x x a Q xn= -] ] ]g g g has a multiple root at x a= then ( )a 0=( )P a P= l 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 infl exion at x a=

Proof

P x x a Q x

P a a a Q a

Q x00

n

n

n$

= -

= -

=

=

] ] ]

] ] ]

]

g g g

g g g

g

Investigation

Use a graphics calculator or graphing computer software to draw graphs with multiple roots.

Examine values close to the roots .(a) Look at the relationship between the degree of the polynomial, the (b) leading coeffi cient and its graph .

Here are some examples of polynomials but you could choose others to examine .

1. P x x x1 3= + -] ] ]g g g 2. P x x x1 3= + -2] ] ]g g g 3. P x x x1 3= - + -3] ] ]g g g 4. P x x x1 3= - + -4] ] ]g g g 5. P x x x1 3= + - 2] ] ]g g g 6. P x x x1 3= + - 3] ] ]g g g 7. P x x x1 3= - + - 4] ] ]g g g 8. P x x x1 3= - + -2 2] ] ]g g g 9. P x x x1 3= - + -2 3] ] ]g g g 10. P x x x1 3= + -3 2] ] ]g g g

Where there is a multiple root, there is always a stationary point (maximum, minimum or point of infl exion).

This means that 0dx

dy= at that point.

If the root is at x a= , then we can write this as ( )a 0= .Pl

See class investigation on pages 686–690.

ch12.indd 701 7/13/09 12:21:00 PM


EXAMPLES

1. Draw a sketch of .P x x x 3 3= - -] ]g g

Solution

Roots of the polynomial equation :P x 0=] g

,x x

x x

x

3 00 3 0

3

3- - =

= - =

=

] g

There is a single root at 0x = so the curve crosses the x -axis at this point.

There is a triple root at 3.x = Since n is odd, there is a point of infl exion at 3.x =

P x x x

x x x xx

3

9 27 27

3

3 2

4 f

= - -

= - - + -

= -

] ]

^

g g

h

Since –x4 is the leading term, P ( x ) has degree 4 and the leading coeffi cient is negative. So as x becomes large (both negative and positive) the value of P x " 3-] g .

30

y

x

( )P x u v v u= +l l l

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

n x a Q x Q x x a

n x a Q x x a Q x

1n n

n n

1

1

$ $ $= - + -

= - + -

-

-

l

l

( )P al ( ) ( ) ( ) ( )n a a Q a a a Q an n1= - + -- l

( ) ( )n Q a Q a0 0

0

n n1$ $ $= +

=

- l

( ) ( )P a P a 0` = =l

There is no need to fully expand the polynomial as we only want to fi nd the leading term.

ch12.indd 702 7/11/09 10:53:51 PM


2. A polynomial has a double root at 5.x = Write an expression for the polynomial .(a) Prove that (b) ( )P P5 5 0= =l] g .

Solution

If (a) P ( x ) has a double root at 5x = , then x 5- 2] g is a factor P x x Q x5So = - 2] ] ]g g g

(b)

P x x Q x

P Q

Q

5

5 5 5 5

0 50

2#

= -

= -

=

=

2

2

] ] ]

] ] ]

]

g g g

g g g

g

To fi nd ( )P 5l , fi rst we differentiate P ( x ) using the product rule.

( )P x u v v u= +l l l

1( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

x Q x Q x x

x Q x x Q x

Q Q

Q Q

2 5 1 5

2 5 5

2 5 5 5 5 5 5

2 0 5 0 50

2

2

2

2

$ $ $

# # #

= - + -

= - + -

= - + -

= +

=

l

l

l

l

3. A monic polynomial has degree 5 and has a double root at a 1 and a triple root at a 2 . Draw a sketch of the polynomial where .a a1 21

Solution

Since P ( x ) is monic and has degree 5, the leading term is x 5 . We could write P x x a x a1 2= - - 32] _ _g i i . Since the polynomial has odd degree and a positive leading coeffi cient, as x becomes a positive large value, P x "3] g and as x becomes a negative large value, P x " 3-] g . The double root at x a1= gives a maximum or minimum turning point and the triple root at x a2= means a point of infl exion. Putting all this information together gives the graph below.

y

xa1 a2

You learned this rule in Chapter 4.

( )P 5l

ch12.indd 703 7/11/09 10:53:55 PM


1. Find the roots of each polynomial equation ( )P x 0= and state if they are multiple roots .

(a) 6 9P x x x2= - +] g (b) P x x x x9 143 2= - +] g (c) 3P x x x3 2= -] g (d) 2 4 8f x x x x3 2= - - +] g (e) P x x x x6 12 83 2= +- -] g (f) 4 5 2A x x x x x4 3 2= - + -] g (g) P x x x x4 24 3 2= - - +] g

x12 9+ (h) Q x x x x8 165 4 3= - +] g (i) P x x x x2 124 3 2= + - +] g

x14 5- (j) 8 36 54 27f x x x x3 2= - + -] g

2. For each graph, state if (i) the leading coeffi cient is

positive or negative and (ii) the degree of the polynomial

is even or odd. y

x

(a)

y

x

(b)

(c) y

x

(d) y

x

y

x

(e)

y

x

(f)

12.5 Exercises

ch12.indd 704 7/10/09 11:59:00 PM


y

x

(g)

y

x

(h)

y

x

(i)

y

x

(j)

3. 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?

4. A polynomial of degree 3 has a triple root at .x 1=

Write down an expression for (a) the polynomial. Is this unique?

If (b) ,P 2 5=] g write the expression for the polynomial.

5. Polynomial P x x x73 2= - +] g x8 16+ has a double root at .x 4=

Show that (a) x 4- 2] g is a factor of P ( x ) .

Write (b) P ( x ) as a product of its factors .

Prove (c) ( ) .4 0=P P4 = l] g

6. Polynomial f x x x74 3= + +] g x x9 27 542 - - has a triple root at 3x = - .

Show that (a) 3x 3+] g is a factor of f ( x ) .

Write (b) f ( x ) as a product of its factors .

Prove (c) ( 3) 0- = .f f3- = l] g

7. A polynomial has a triple root at x k= and degree n .

Write an expression for the (a) polynomial .

Prove that (b) ( ) ( ) .P k P k 0= =l

8. Draw an example of a polynomial with leading term

(a) x 3 (b) 2x5- 3 (c) x 2 (d) x4- (e) 2x3-

9. Draw an example of a polynomial with a double root at 2x = and a leading term of 2 x 3 .

10. Draw an example of a polynomial with a double root at x 1= - and leading term x3- .

11. Sketch an example of a polynomial with a double root at 2x = and a leading term of x 4 .

12. Draw an example of a polynomial with a double root at 3x = - and leading term .x6

ch12.indd 705 7/11/09 9:37:27 PM


13. A polynomial has a triple root at x 1= and it has a leading term of x 3 . Draw an example of a graph showing this information.

14. Given a polynomial with a triple root at 0x = and leading term ,x4- sketch a polynomial on a number plane that fi ts this information.

15. If a polynomial has a triple root at 2x = - and a leading term of x 8 sketch a polynomial fi tting this information.

16. A polynomial has a triple root at 4x = and its leading term is .x4 3- Show this on a number plane.

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 12-

18. A polynomial with leading term x8- has a triple root at 2x = - . Show by a sketch that the polynomial has at least one other root in the domain .x 22-

19. A polynomial has a double root at 2x = and a double root at .x 3= - Its leading term is 2 x 5 . By drawing a sketch, show that the polynomial has another root in the domain .x 22

20. Show that a polynomial with leading term x3- and a double root at x 1= has another root at a point where .x 12

Roots and Coeffi cients of Polynomial Equations

In Chapter 10, you studied the relationship between the roots and coeffi cients 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 bx c 02 + + = can be written in monic form as

0x ab x a

c2 + + =

If the quadratic equation has roots a and ,b then the equation can be written in monic form as ( ) ( )

( )

x x

x x x

x x

0

0

0

2

2

a b

b a ab

a b ab

- - =

- - + =

- + + =

i.e. ( )x ab x a

c x x2 2/ a b ab+ + - + +

ch12.indd 706 7/10/09 11:59:02 PM


Cubic equation

The cubic equation 0ax bx cx d3 2+ + + = can be written in monic form as

0.x ab x a

c x ad3 2+ + + =

If the cubic equation has roots ,a b and c then the equation can be written in monic form as

( ) ( ) ( )

( ) ( )

( ) ( )

x x x

x x x x

x x x x x x x

x x x

0

0

0

0

2

3 2 2 2

3 2

a b c

b a ab c

c b bc a ac ab abc

a b c ab bc ac abc

- - - =

- - + - =

- - + - + + - =

- + + + + + - =

( ) ( )x ab x a

c x ad x x x3 2 3 2/ a b c ab bc ac abc+ + + - + + + + + -

This gives the results below:

For the quadratic equation 0:ax bx c2 + + = Sum of roots: •

ab

a b+ = -

Product of roots: •

ac

ab =

For the cubic equation :ax bx cx d 03 2+ + + =

Sum of roots 1 at a time: •

ab

a b c+ + = -


ac

ab ac bc+ + =

Product of roots (sum of roots 3 at a time) •

ad

abc = -


Quartic equation

The quartic equation 0ax bx cx dx e4 3 2+ + + + = can be written in monic form

as 0.x ab x a

c x ad x a

e4 3 2+ + + + =

ch12.indd 707 7/10/09 11:59:03 PM


If the quartic equation has roots , ,a b c and d then the equation can be written in monic form as

0=

( ) ( ) ( ) ( )

[ ( ) ( ) ] ( )

( ) ( )

( ) ( )

( ) ( )( )

x x x x

x x x x

x x x x

x x x

x x xx

0

0

0

3 2

4 3 3 2

2

4 3 2

a b c d

a b c ab bc ac abc d

d a b c d a b c

ab bc ac d ab bc ac abc abcd

a b c d ad db dc ab bc ac

abd bdc adc abc abcd

- - - - =

- + + + + + - - =

- - + + + + + +

+ + - + + - +

- + + + + + + + + + -

+ + + + =

x ab x a

c x ad x a

e4 3 2` + + + +

( ) ( )

( )x x x

x

4 3 2/ a b c d ad db dc ab bc ac

abd bdc adc abc abcd

- + + + + + + + + +

- + + + +


For the quartic equation 0:ax bx cx dx e4 3 2+ + + + = Sum of roots 1 at a time: •

ab

a b c d+ + + = -


ac

ab ac ad bc bd cd+ + + + + =


ad

abc abd acd bcd+ + + = -

Product of roots (sum of roots 4 at a time): •

ae

abcd =

This pattern extends to polynomials of any degree.

Class Investigation

Can you fi nd results for sums and products of roots for equations of degree 5, 6 and so on?

ch12.indd 708 7/10/09 11:59:03 PM


EXAMPLES

1. If , ,a b c are the roots of ,x x x2 5 1 03 2- + - = fi nd

(a ) ( ) 2a b c+ +

(b) ( ) ( ) ( )1 1 1a b c+ + +

(c) .1 1 1a b c+ +

Solution

( )

( )

ab

ac

ad

25

25

21

21

21

a b c

ab ac bc

abc

+ + = -

= --

=

+ + =

=

= -

= --

=

(a) ( )25 6

412

2

a b c+ + = =c m

(b) ( ) ( ) ( )1 1 1a b c+ + + ( ) ( )

( ) ( )

1 1

1

1

21

21

25 1

421

a bc b c

abc ab ac a bc b c

abc ab ac bc a b c

= + + + +

= + + + + + + +

= + + + + + + +

= + + +

=

1 1 1

2121

1

a b c abc

bc ac ab+ + =

+ +

=

=

(c)

CONTINUED

ch12.indd 709 7/11/09 9:37:30 PM


2. If one root of 2 3 0x x x3 2- + - = is 4, fi nd the sum and product of the other two roots.

Solution

Roots are , ,a b c where, say, 4c = .

( )

ab

ad

4 1

3

4 3

43

`

`

a b c

a b

a b

abc

ab

ab

+ + = -

+ + =

+ = -

= -

=

=

3. Solve ,x x x12 32 15 9 03 3+ + - = given that 2 roots are equal.

Solution

Let the roots be ,a a and .b

( )

( )

( )

ab

ac

ad

21232 1

21215 2

129 3

2

2

`

`

`

a a b

a b

aa ab ab

a ab

aab

a b

+ + = -

+ = -

+ + =

+ =

= -

=

From ( ):

( )

Substitute in ( ):

( ) ( )

1

1232 2 4

2

21232 2

1215

12 241232 2 15

12 64 48 15

0 36 64 152 3 18 5

2 3 0 18 5 0

2 3 18 5

121

185

2

2

2 2

2

b a

a a a

a a a

a a a

a a

a a

a a

a a

a a

= - -

+ - - =

+ - - =

- - =

= + +

= + +

+ = + =

= - = -

= - =-

c

c

m

m

ch12.indd 710 7/10/09 11:59:04 PM


Substitute in (4):

1 : 2 121

: 2185

2

1232

1

1232

1

21

3

185

9

a b

a b

= - = - - -

=

= - = - - -

= -

c

c

m

m

Substitute in (3):

, :

, :

121

31 1

21

31

129

43

43

185 2

91

185 2

91

129

2

2

a b

a b

= - = - =

=

= - = - - - =

c c

c c

m m

m m

This is impossible as LHS is negative and RHS is positive.

the roots are 121

- and 31

1. Given that a and b are the roots of the equation, fi nd (i) a b+ and (ii) ab for the following quadratic equations.

(a) 2 8 0x x2 - + = (b) 3 6 2 0x x2 + - = (c) x x7 1 02 + + = (d) x x4 9 12 02 - - = (e) 5 15 0x x2 + =

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) 2 8 0x x x3 2+ - + = (b) 3 5 2 0x x x3 2- + - = (c) 2 6 2 0x x x3 2- + + = (d) x x3 11 03 2- - - = (e) 7 3 0x x3 + - =

3. For the following quartic equations, where , ,a b c and d are the roots of the equation, fi nd

(i) ,a b c d+ + + (ii) ,ab ac ad bc bd cd+ + + + + (iii) abc abd acd bcd+ + + and (iv) abcd

(a) 2 5 0x x x x4 3 2+ - - + = (b) 3 2 7 0x x x x4 3 2- - + - = (c) 3 2 4 0x x x x4 3 2- + + - + = (d) 2 2 4 3 2 0x x x x4 3 2- - + - = (e) 2 12 7 0x x4 3- + =

4. If a and b are the roots of 5 5 0,x x2 - - = fi nd

(a) a b+

(b) ab

(c) 1 1a b+

(d) 2 2a b+

(e) 3 3a b+

12.6 Exercises

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5. If ,a b and c are the roots of 2 5 3 0,x x x3 2+ - - = fi nd

(a) abc (b) ab ac bc+ + (c) a b c+ +

(d) 1 1 1a b c+ +

(e) ( ) ( ) ( )1 1 1a b c+ + +

6. If , ,a b c and d are the roots of 2 5 3 0,x x x4 3- + - = fi nd

(a) abcd (b) abc abd acd bcd+ + +

(c) 1 1 1 1a b c d+ + +

7. One root of 3 2 0x x k2 - + - = is 4.- Find the value of k .

8. One root of 5 21 0x x x3 2- - + = is 3. Find the sum a b+ and the product ab of the other two roots.

9. Given ( ) ,P x x x x2 7 4 13 2= - + + if the equation ( ) 0P x = has a root at ,x 1= fi nd the sum and product of its other roots.

10. Find the value(s) of k if the quadratic equation ( )x k x k2 1 02 - + + + = has

equal roots (a) one root equal to 5 (b) consecutive roots (c) one root double the other (d) reciprocal roots (e)

11. Two roots of 15 7 0x mx x3 2+ + - = are equal and rational. Find m .

12. Two roots of 5 0x ax bx3 2+ + - = 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 x x2 7 64 3= - + -

If (b) ,a b and c are the other 3 zeros, fi nd the value of a b c+ + and .abc

14. If 2x = is a double root of 2 8 16 0,ax x x4 3- - + = fi nd the value of a and the sum of the other two roots.

15. Two of the roots of 4 0x px qx3 2- - - = are 3 and 5.

Find the other root. (a) Find (b) p and q .

16. The product of two of the roots of 2 18 5 0x x x4 3+ - - = is 5.- Find the product of the other two roots.

17. The sum of two of the roots of x x x x7 14 1 04 3 2+ + + - = is 4. Find the sum of the other two roots.

18. Find the roots of 8 20 6 9 0,x x x3 2- + + = given that two of the roots are equal.

19. Solve x x x12 4 3 1 03 3- - + = if the sum of two of its roots is 0.

20. Solve 6 5 24 15 18 0x x x x4 3 2+ - - + = if the sum of two of its roots is zero.

ch12.indd 712 7/25/09 11:41:24 AM


Test Yourself 12 1. Write ( )p x x x x x4 14 36 454 3 2= + - - +

as a product of its factors.

2. If ,a b and c are the roots of 3 9 0,x x x3 2- + - = fi nd

(a) a b c+ +

(b) abc

(c) ab ac bc+ +

(d) 1 1 1a b c+ +

3. A monic polynomial ( )P x of degree 3 has zeros 2,1- and 6. Write down the polynomial.

4. (a) Divide ( )P x x x x x19 49 304 3 2= + - - - by .x x2 152 - -

Hence, write (b) ( )P x as a product of its factors.

5. For the polynomial ( ) ,P x x x x2 33 2= + - fi nd

the degree (a) the coeffi cient of (b) x the zeros (c) the leading term .(d)

6. Sketch ( ) ( ) ( )f x x x2 3 2= - + showing the intercepts.

7. If 3 48 60 0ax x x x4 3 2+ - + = has a double root at 2,x = fi nd

the value of (a) a the sum of the other two roots .(b)

8. Show that 7x + is not a factor of 7 5 4.x x x3 2- + -

9. If the sum of two roots of x x x x2 8 18 9 04 3 2+ - - - = is 0, fi nd the roots of the equation.

10. The polynomial ( )f x ax bx c2= + + has zeros 4 and 5, and ( ) .f 1 60- = Evaluate a , b and c .

11. Find the x - and y -intercepts of the curve .y x x x3 10 243 2= - - +

12. Divide ( ) 3 7 8 5p x x x x5 3 2= - + - by 2,x - and write ( )p x in the form ( ) ( 2) ( ) ( ) .p x x a x b x= - +

13. Solve cos cos cosx x x2 03 2+ - = for .x0 360c c# #

14. When 8 5 9x kx3 - + is divided by 2,x - the remainder is .1- Evaluate k .

15. Find the zeros of ( ) 9 20.g x x x2= - + -

16. Sketch ( ) 2 ( 3)( 5),P x x x x= - + showing intercepts.

17. Find the value of k if the remainder is 4- when 2 3x x x k3 2+ - + is divided by 2.x -

18. The sum of 2 roots of 7 5 3 0x x x x4 3 2- + - + = 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 ( ) .a 0=( )P a P= l

20. Show that 5x - is a factor of ( ) .f x x x x6 12 353 2= - + -

21. (a) Show that 5x - is a factor of f x 3 2x x x7 5 75= - - +] g .

Show that (b) ( )f 5 5 0= .f= l] g What can you say about the root (c)

at 5x = ? Write (d) f ( x ) as a product of its factors.

22. The leading term of a polynomial is 3x3 and there is a double root at 3x = . Draw an example of a graph of the polynomial.

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23. A polynomial P ( x ) has a triple root at 6x = - .

Write an expression for (a) P ( x ) . If (b) P ( x ) has leading coeffi cient 3 and

degree 4, draw a sketch showing this information.

24. Draw an example of a polynomial with leading term .x3 5

25. If P x ax bx cx d3 2= + + +] g has a remainder of 8 when divided by , ,x P P1 2 17 1 4- = - = -] ]g g and 0 5P =] g , evaluate a , b , c and d .

1. Write ( )P x x x x x x2 2 15 4 3 2= + + - - - as a product of its factors.

2. A polynomial ( ) ( ) ( ) .P x x b Q x7= -

Show that (a) ( ) .b 0=( )P b P= l

Hence fi nd (b) a and b, if ( )x 1 7- is a factor of

( ).

P x x x ax x x bxx

3 31

7 6 5 4 3 2= + + + + + -

+

3. Solve tan tan tan tan3 3 04 3 2i i i i- - + = for .0 360c c# #i

4. (a) Find the equation of the tangent to the curve y x3= at the point where .x 1=

Find the point where this tangent (b) cuts the curve again.

5. (a) Find the remainder when ( ) 2 7 3 9p x x x ax x4 3 2= - + + - is

divided by .x2 1-

If the remainder, when (b) ( )p x is divided by 2,x + is 17, fi nd the value of a.

6. If ,a b and c are roots of the cubic equation 2 8 6 0,x x x3 2+ - + = fi nd

abc(a) 2 2 2a b c+ +(b)

7. Solve sin sin4 3 1 03 i i- - = for .0 360c c# #i

8. Find the value of a if ( ) ( )x x1 2+ - is a factor of 2 2.x x ax3 2- + -

9. Prove that if x a- is a factor of polynomial ( ),P x then ( ) 0.P a =

10. Find the points of intersection between the polynomial P x x x x5 4 13 2= + + -] g and the line 3 4 0x y+ + = .

11. Write down an example of a polynomial with the graph below.

y

x2-1

12. Sketch an example of a polynomial with a double root at x a1= and a double root at x a2= , if the polynomial is monic and has even degree .a a2 12_ i

Challenge Exercise 12

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