Continuous function: A function is continuous over an interval if it has no break in its graph. For every x value on the graph the limit exists and equals the function value

Halving the interval: Taking an x-value halfway between 2 other values and determining the sign of the y-value in order to find an approximate solution to an equation

Interval: A section of a line or curve over a restricted domain including the endpoints

Iteration: Starting with an initial value, often a guess, and using a process over and over again to produce a closer approximation

Newton’s method of approximation: An approximation method to find a solution to an equation. If x a= is close to the root of a polynomial, then the x-intercept of the tangent at that point will generally be closer

Polynomial: A function of the form ( )P x a a x

0 1= + + a x a x…

2

2

n

n+ + where , ,a a a…0 1 n

are real numbers and n is a positive integer or zero

Root of a polynomial: The solution of a polynomial equation. Graphically, it is where the polynomial crosses the x-axis

TERMINOLOGY

Polynomials 2

9

447Chapter 9 Polynomials 2

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-’?

INTRODUCTION

POLYNOMIALS ARE AN IMPORTANT part of algebra and are used in many branches of mathematics. You studied polynomials in the Preliminary Course.

In this chapter you will study the estimation of roots of polynomials by using the methods of halving the interval and Newton’s method.

Estimation of Roots

The roots of a polynomial equation …a a x a x a x 0nn

0 1 22+ + + + = are the

x-intercepts of the graph ( ) . . .P x a a x a x a xnn

0 1 22= + + + +

y

x

P(x)= a0+ a1x+ a2x2+…+ anxn

P(x)= 0 so a0+ a1x+ a2x2+…+ anxn= 0

In some cases, we can find the roots of the equation by using algebra.

448 Maths In Focus Mathematics Extension 1 HSC Course

ExamplEs

1. Find the exact roots of the quadratic equation .x x3 1 02 − + =

Solution

( )( ) ( ) ( ) ( )

±

±

±

±

xa

b b ac2

4

2 13 3 4 1 1

23 9 4

23 5

2

2

= − −

=− − − −

= −

=

Note: The roots are the x-intercepts of the graph of the parabola .y x x3 12= − +

3+ 52

y

x

23− 5

2. For the polynomial ( )P x x x x2 5 63 2= − − +show that (a) 1x − is a factor of the polynomial.write the polynomial as a product of its factors.(b) find the roots of the polynomial equation (c) .x x x2 5 6 03 2− − + =

Solution

If (a) x 1− is a factor of the polynomial, then ( ) .P 1 0=

3 2

( )

( )

x

1

3

( ) ( )

P x x x

P

2 5 6

1 2 1 5 1 60

2= − − += − − +=

So 1x − is a factor.

Remember from the Preliminary Course that x a− is a factor of P(x) if P(a) = 0.

449Chapter 9 Polynomials 2

Divide the polynomial by (b) x 1− to find other factors.

x x x x

x x

x x

x xx

x

1 2 5 6

5

6 6

6 6

0

2

3 2

3 2

2

2

− − − +−− −− +

− +− +

x x 6− −g

So ( )x ( ) ( )( ) ( ) ( )

P x x xx x x

1 61 3 2

2= − − −= − − +

(c) ( ) ( ) ( )

x x xx x x

2 5 6 01 3 2 0

3 2− − + =− − + =

, ,

, ,

x x x

x x x

1 0 3 0 2 0

1 3 2

− = − = + == = = −

Note: The roots are the x-intercepts of the graph of the polynomial ( )xP x x x2 5 63 2= − − +

y

x-2 1 3

It is not always possible to find the roots of a polynomial equation by algebraic methods. We can estimate the roots by looking at the graph.

ExamplE

Use stationary points to sketch the curve y x x x2 3 12 53 2= + − + and use the graph to give a rough estimate of the roots of the equation

x x x2 3 12 5 03 2+ − + =

Solution

y x x x

dx

dyx x

2 3 12 5

6 6 12

3 2

2

= + − +

= + −

You learned how to do this in the Preliminary Course.

CONTINUED

450 Maths In Focus Mathematics Extension 1 HSC Course

For stationary points .dx

dy0=

( ) ( )

x x

x xx x

6 6 12 0

2 02 1 0

2

2

+ − =+ − =

+ − =,

,

x x

x x

2 0 1 0

2 1

+ = − == − =

When x 2= −3 2( ) ( ) ( )y 2 2 3 2 12 2 5

25= − + − − − +=

When 1x =3 2( ) ( ) ( )y 2 1 3 1 12 1 5

2= + − += −

So ( , )2 25− and ( , )1 2− are stationary points.

dx

d yx12 6

2

2

= +

At ( , )2 25−

( )dx

d y12 2 6

18

2

2

= − +

= −0< (concave downwards)

So ( , )2 25− is a maximum turning point.

At ( , )1 2−

( )dx

d y12 1 6

18

2

2

= +

=0> (concave upwards)

So ( , )1 2− is a minimum turning point.

For x-intercept, 0y =x x x0 2 3 12 53 2= + − +

This will not factorise so we cannot find the x-intercepts.

For y-intercept, 0x =3 2( ) ( ) ( )y 2 0 3 0 12 0 5

5= + − +=

There are 3 x-intercepts, so

5

(1, -2)

(-2, 25)y

x

3 roots of the equation

.x x x2 3 12 5 03 2+ − + =One root is where ,x 2< − say around .x 3= − Another root is between 0x = and ,x 1= say . .x 0 2= The third root is where ,x 1> say .x 3=

451Chapter 9 Polynomials 2

Halving the interval

If ( )f x is continuous for ≤ ≤a x b and ( )f a and ( )f b have opposite signs, then there is at least one root of ( )f x 0= in that interval.

There are two iterative methods of estimating roots of polynomial equations that you will study in this course. These are called halving the interval and Newton’s method of approximation. In these methods, we guess a solution and then use a process over and over to produce closer approximations.

These methods can also be used to find roots of functions that are not polynomials, as long as the function is continuous over the interval.

DID YOU KNOW?

Niels Abel (1802–29), a Norwegian mathematician, proved that it is impossible to find a general method to solve quintic equations (equations of degree 5). He was 22 years old when he made this discovery. Abel established many theorems in his short life, especially in the area of group theory, which is one of the fundamentals of abstract algebra. He died at the age of 26.

452 Maths In Focus Mathematics Extension 1 HSC Course

If ( ) 0f a < and ( ) 0f b > , then there are three possibilities:

1. f a b2

0+ =c m means a b2+ is a root of the equation

2. f a b2

0<+c m means the root lies between x a b

2= + and

x b=

3. a bf2

0>+c m means the root lies between x a= and

a bx2+=

This method is also called bisection. We can find an approximation to the root between a and b by halving

this interval.

453Chapter 9 Polynomials 2

If we halve the interval several times, the approximation to the root will usually, but not always, become more accurate.

ExamplEs

1.

Show that a root of (a) x x x3 9 1 03 2− − + = lies between x 4= and .x 5=By halving the interval, show that the root lies between 4.75 and 4.875.(b)

Solution

(a) ( ) ( ) ( )

( ) ( ) ( )

f

f

4 4 3 4 9 4 119

5 5 3 5 9 5 16

3 2

3 2

= − − += −

= − − +=

Since (4) 0f < and (5) 0,f > the root lies between 4 and 5.

(b) ( . )

. ( . ) ( . ).

f f2

4 5 4 5

4 5 3 4 5 9 4 5 19 125

3 2

+ =

= − − += −

c m

Since (4.5) 0,f < the root lies between 4.5 and 5.

CONTINUED

454 Maths In Focus Mathematics Extension 1 HSC Course

2

. ( . )

. ( . ) ( . ).

f f2

4 5 5 4 75

4 75 3 4 75 9 4 75 12 265625

3

+ =

= − − += −

c m

Since ( . ) ,f 4 75 0< the root lies between 4.75 and 5.

. ( . )

. ( . ) ( . )

.

f f2

4 75 5 4 875

4 875 3 4 875 9 4 875 11 686

3 2

+ =

= − − +=

c m

Since ( . ) ,f 4 875 0> the root lies between 4.75 and 4.875.

Note: Since f (4.875) is closer to 0 than f (4.75), .x 4 875= is a better approximation than 4.75.

2.

Show that (a) x x x3 4 12 1 04 3 2+ − − = has a root between 3x = − and 2.x = −

Use the method of halving the interval to show that the root lies (b) between 3− and . .2 75−

Solution

(a) 4 2( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

f

f

3 3 3 4 3 12 3 126

2 3 2 4 2 12 2 133

3

4 3 2

− = − + − − − −=

− = − + − − − −= −

455Chapter 9 Polynomials 2

Since ( )f 3 0>− and ( ) ,f 2 0<− the root lies between 3− and 2.−

(b) ( . )

( . ) ( . ) ( . ).

f f2

3 2 2 5

3 2 5 4 2 5 12 2 5 121 3125

4 3 2

− + − = −

= − + − − − −= −

c m

Since ( . ) ,f 2 5 0<− the root lies between 3− and 2.5.−

. ( . )

( . ) ( . ) ( . ).

f f2

3 2 5 2 75

3 2 75 4 2 75 12 2 75 13 363

4 3 2

− + − = −

= − + − − − −= −

c m

Since ( . ) ,f 2 75 0<− the root lies between 3− and 2.75.−

Note: ( . )f 2 75− is closer to zero than ( ) .f 3− So 2.75− is a closer approximation to the root than 3.−

This does not always happen. Sometimes the first or second value of ( )f x found can be closer to the root than subsequent values.

456 Maths In Focus Mathematics Extension 1 HSC Course

Computer application

The method of approximation by halving is tedious and can easily be done on a computer. Some computer programs already have this method built in. Otherwise,

you could put the formula into a spreadsheet.

1. Show that there is at least one root between the x-values given for each function.(a) ( )f x x x2 52= − − between

3x = and 4x =(b) ( )f x x x7 2= + − between 3x = and 4x =(c) ( )p x x x7 2= + − between x 3= − and 2x = −(d) ( )P x x x x2 3 72 13 2= − − + between 0x = and 1x =(e) ( )f x x x5 24 2= − + between

2x = and .x 3=

2. (a) Show ( )f x x x2 5 22= + − has a root between 0x = and .x 1=

By halving the interval once, (b) show that the root lies between

0x = and 0.5.x =

3. Show that there is no root between 0x = and 1x = on the curve ( ) .f x x x x2 3 13 2= − + +

4. (a) Show ( )P x x x4 153 2= + − 21 5x − has a

root between 1x = and .x 2=By halving the interval twice, (b)

show that P(x) has a zero of 1.25.

5. (a) Show ( )f x x x 32= + − has a root between 1x = and .x 2=

By halving the interval once, (b) show that the root lies between

1x = and 1.5.x =

6. The function ( )f x x x5 12= + + has a root between 5x = − and 4.x = − By halving the interval, show that the root lies between 5x = − and 4.5.x = − Which is the better approximation?

7. Find an approximation to the root of the function ( )f x x 27= − that lies between 1x = and 2,x = by halving the interval once.

8. The curve y x x 72= − − has a root between 3x = and 4.x = By halving the interval twice, find an approximation to the root.

9. The function ( )f x x x4 13 2= + + has a root between 5x = − and

4.x = − Halve the interval twice and find an approximation to the root.

10. (a) Sketch the function ( ) .f x x x x2 9 12 13 2= + + +

How many zeros does (b) ( )f x have? Between which x-values do they lie?

By halving the interval twice, (c) find approximations to these zeros.

11. (a) Show that sine x 3x = + has a root between 1x = and .x 2=

Use the method of halving (b) the interval three times to find a closer approximation to the root.

9.1 Exercises

457Chapter 9 Polynomials 2

12. For the polynomial ( )P x x 123= −show that when (a) ( ) ,P x 0=

x 12= 3

show that a root of (b) ( )P x 0= lies between 2x = and 3x =

use the method of halving (c) the interval twice to find an approximation to .123

13. (a) Show that a root of

sin xx5

0− = lies between 2x =

and .x 3=Use the method of halving (b)

the interval twice to find an estimate of the root.

14. (a) Show that a root of ( )ln x e3 0x+ − = lies between 0.1x = and . .x 0 2=By halving the interval, (b)

show that the root lies between 0.125x = and . .x 0 15=

15. A root of ( )P x x2x 3= − lies between 1x = and 2.x = Show that the root lies between

1.25x = and 1.375x = by using the method of halving the interval.

Newton’s method of approximation

Newton’s method is a different way of approximating the root of a polynomial equation. It generally gives a more accurate approximation than the method of halving the interval, as well as taking fewer steps to get this approximation.

Sketching ( ),y f x= a continuous function, shows how Newton’s method works.

If x a= is close to the root of the equation ( ) 0,f x = then the x-intercept (a1) of the tangent at a is usually closer to the root.

The function only needs to be continuous in the

interval near the root.

( )

( ),a a

a

f a1 = −

fl where x a= is close to the root

458 Maths In Focus Mathematics Extension 1 HSC Course

Proof

Let [ , ( )]P a f a= be a point on the curve ( )y f x= and let point R be the x-intercept of the tangent at P.The gradient of the tangent to the curve is given by ( )xfl\ the gradient of the tangent at P is given by ( )aflThe equation of the tangent at P is given by

( ) ( )a x a−( )

( )

y y m x x

y f a f1 1− = −

− = l

At R, 0y =i.e.

( )a

( )a x

( )( )

( )

a x a

a x af

−−

( )

( )

( ) ( )

( )

( ) ( )

( )

( )

( )

( )

( )

( )

f a f

f a f

a f a f

a

a f ax

a

a

a

f ax

aa

f ax

0 − =− =− =−

=

− =

− =

af

f

f

af

af

f

f

l

l l

l l

l

l

l

l

l

l

the x-intercept a1 of the tangent at P is given by

( )

( )a a

f a

f a1 = −

l

Several applications of Newton’s method will often give a closer approximation.

You could use a computer for these calculations.

We often halve the interval to get the first approximation of x a.=

ExamplEs

1. Find an approximation to the root of x x 1 03 + − = by using Newton’s method once and starting with an approximation of 0.5.x =

\

459Chapter 9 Polynomials 2

CONTINUED

(0.714) 0.079,Zf so x 0.714= is a good

approximation.

Sketch (x)y f= by finding its axis of symmetry, or using a

table of values.

Solution

( . ) . ..

( )

( . ) ( . ).

( )

.( . )

...

.

f

x x

a af a

f

0 5 0 5 0 5 10 375

3 1

0 5 3 0 5 11 75

0 50 5

0 51 750 375

0 714

3

2

2

1

= + −= −

= += +=

= −

= −

= − −

=

( )

f

a

f

( . )

f

0 5f

l

l

l

l

2. Sketch (a) ( )f x x x4 12= − − and show on your sketch that one root of

x x4 1 02 − − = lies between 4x = and 5.x =By choosing an approximation of (b) 4.5,x = use Newton’s method once

to find an approximation to this root.

Solution

(a)

460 Maths In Focus Mathematics Extension 1 HSC Course

(b) ( . ) . ( . ).

( )

( . ) ( . )

f

x x

4 5 4 5 4 4 5 11 25

2 4

4 5 2 4 5 45

2= − −== −= −=

f

f

l

l

( . )4 5f

( )a

( )

.( . )

. .

.

a af

f a

f4 5

4 5

4 55

1 25

4 25

1 = −

= −

= −

=

l

l

So an approximation to the root is 4.25.x =

3. Find an approximation to the root of cos x x3

= near . .x 1 1=

Solution

cos x x3

=

i.e. cos x x3

0− =

Let ( ) cosf x x x3

= −

Then ( ) sinx x= − −f31

l

Use 1.1x = as the first approximation.

.1 17( )1 1

( . ) . .

.

( . ) .

.

( )

( )

.( . )

cos

sin

f

a af a

f a

f

f

1 1 1 13

1 1

0 0869

1 1 1 131

1 2245

1 11 1

1

Z

= −

=

= − −

= −

= −

= −

f

.

l

l

l

f (4.25) 0.0625,= so x 4.25= is a good approximation.

1.17 is a much closer approximation to the root than 1.1.

Newton’s method doesn’t work in some cases.

The approximate root of

3cos x =x

was found

graphically to be x 1.1= in Chapter 5.

461Chapter 9 Polynomials 2

ExamplE

Find an approximation to the root of y x x 13= + − by using Newton’s method once and starting with an approximation of 0.2.x =

Solution

( . ) . ..

( )

( . ) ( . ).

f

x x

0 2 0 2 0 2 10 792

3 1

0 2 3 0 2 11 12

3

2

2

= + −= −

= += +=

f

f

l

l

( )

( )

.( . )

...

.

a af a

f a

f

f0 2

0 2

0 21 120 792

0 907

1 = −

= −

= − −

=

( . )0 2

l

l

This is not a close approximation to the root.

The root lies between x 0= and x 1.= The

approximation of x 0.2= is not a good choice as it is too far from the root.

f (0.907) 0.6 ,5Z which is not very close to 0.

Class Investigation

Here are some examples where the approximation of a root either is not very close to the root, or cannot be found. Discuss these examples.

1.

In this case, the approximation of a1 will be further away from the root than a. Can you see why?

462 Maths In Focus Mathematics Extension 1 HSC Course

1. The polynomial equation x x2 9 02 − − = has a root near

2.5.x = Find an approximation to the root, correct to 2 decimal places, by using 1 application of Newton’s method.

2. The equation ( ) 0f x = where ( )f x x x3 93 2= − + has a zero near

1.5.x = − By using 1 application of Newton’s method, find an approximation to the root, correct to 2 significant figures.

3. The polynomial equation x x x4 2 03 2− + − = has a root

near 0.6.x = Use 2 applications of Newton’s method to find an approximation to the root, correct to 3 decimal places.

4. (a) Show that the polynomial equation x x x2 6 3 03 2− − + = has a root between 0x = and .x 1=

Use Newton’s method (b) with 1 application to find an approximation to the root, correct to 3 decimal places, using

0.5x = as an approximation to the root.

5. (a) Show that 2x x 1 0+ − =1

has a root between 0x = and .x 1=

Use 1 application of Newton’s (b) method to find an approximation of the root to 3 decimal places.

9.2 Exercises

2.

In this case, a1 cannot be found.

3.

The x-intercept a1 is too far away from the root in this case.

Can you find any more examples where Newton’s method doesn’t work?

463Chapter 9 Polynomials 2

6. (a) Show that the equation x x x2 5 03 2− + + = has a root between 2x = − and .x 1= −

Use the method of halving (b) the interval twice to find an approximation to this root of the equation.

Use Newton’s method once (c) to find an approximation to this root of the equation, correct to 2 decimal places, starting with

.x 1 5= − .

7. (a) Show that the equation x x2 2 02 + − = has a root between

0x = and .x 1=Use the method of halving (b)

the interval twice to find an approximation to this root of the equation.

Use Newton’s method once (c) to find an approximation to this root of the equation, correct to 1 decimal place.

Use the quadratic formula to (d) find an approximation to this root of the equation, correct to 2 decimal places.

8. (a) For the polynomial ( ) ,P x x 93= − show that the

equation ( )P x 0= can be written as .x 9= 3

Between which 2 integers (b) does 93 fall?

By using Newton’s method (c) once, find an approximation to

,93 correct to 2 decimal places.

9. Use Newton’s method with 1 application to find, correct to 2 decimal places,(a) 53

(b) 4005

(c) 554

(d) 507

10. The curve 5( )y x x2 3= − − has a root between 2x = and

3.x = Use Newton’s method with 1 application to find an approximation to the root, correct to 2 decimal places. Is this a good approximation?

11. The equation log x x 2e3= − has

a root between 1x = and 2.x = Use ( ) logf x x x 2e

3= − + and Newton’s method once to find an approximation to the root, correct to 2 decimal places.

12. The root of the equation e x 0x 3− = lies between x 1= and

.x 2= Use Newton’s method once to find an approximation to the root, correct to 2 decimal places.

13. Use a first approximation of 0.6x = with 1 application of

Newton’s method to solve ,a x xt n = correct to 2 decimal

places.

14. (a) Find two integers between which the root of

( 3) 1ln x x+ = − lies.Use Newton’s method once to (b)

find an approximation, correct to 2 decimal places.

15. Use 0.5x = to find an approximation to the root of

,cos x x= correct to 2 decimal places, using 2 applications of Newton’s method.

464 Maths In Focus Mathematics Extension 1 HSC Course

Test Yourself 91. The polynomial equation x x3 2 3 02 − − =

has a root near 1x = . Use two applications of Newton’s method to find a closer approximation to the root, correct to 2 decimal places.

2. Use one application of Newton’s method to find an approximation to the root of ,e x 0x 3− = starting with a first approximation at 2.x =

3. (a) Show that there is a root of ( )f x x x3 54 2= − − between 2x = and

.x 3=Using (b) 2x = as a first approximation,

use Newton’s method to find a closer approximation.

4. (a) Show that the function ( )f x x x5 73= − + has a root between

3x = − and .x 2= −By halving the interval twice, find (b)

the closest approximation to the root.

5. (a) Sketch the curve y x x2 9 73 2= − + showing any stationary points.

Find the (b) x-intercepts on the curve, using Newton’s method where necessary, to find an approximation to 1 decimal place.

6. Find an approximation to the root of x x3 7 03 + − = by halving the interval 3 times.

7. Solve log x x 2e3= − by using two

applications of Newton’s method, starting with a first approximation of 1.x =

8. (a) Use 2x = as a first approximation to find the solution to ,x 105 = using two applications of Newton’s method (answer correct to 2 decimal places).

Find the solution of (b) x 105 = directly.

9. (a) Show that a root of x x3 1 03 − + = lies between 0x = and .x 1=

Explain why Newton’s method will (b) not work with a first approximation of .x 1=

Use Newton’s method with (c) 0x = as a first approximation to find the root of the equation.

Halve the interval twice to find an (d) approximation to the root.

10. A root of sin xx3

= lies between 2x = and

3.x =Halve the interval twice to find an (a)

approximation to the root.Use one application of Newton’s (b)

method to find the root, starting with 2x = as a first approximation.

465Chapter 9 Polynomials 2

1. (a) Sketch the curve .y x x x6 8 14 2= − + +Estimate the roots of equation (b)

,x x x6 8 1 04 2− + + = by choosing appropriate approximations for the roots and using 1 application of Newton’s method.

One of the roots is a poor estimation. (c) Which one is it?

By halving the interval twice, find (d) the best estimate of this root.

2. Given ( ) ,P x x a3= − the equation ( )P x 0= has a root near x b= where .≠b 0 Show that an approximation to the root is

given by b

b ax3

23

3

1

+= .

3. (a) Show that a root of sinx

x3

0−

= lies

between 6x = and . .x 6 5=Use the method of halving the (b)

interval to show that this root lies between 6.25x = and . .x 6 375=

Use Newton’s method to find an (c) approximation of this root correct to 3 decimal places using a first approximation of 6.25.

Challenge Exercise 9