Polynomials

Expressions like 3x4 + 2x3 – 6x2 + 11 and m6 – 4m2 +3 are called polynomials.

(5x – 2)(2x+3) is also a polynomial as it can be written 10x2 + 11x - 6.

The degree of the polynomial is the value of the highest power.

3x4 + 2x3 – 6x2 + 11 is a polynomial of degree 4.

m6 – 4m2 + 3 is a polynomial of degree 6.

In the polynomial 3x4 + 2x3 – 6x2 + 11, the coefficient of x4 is 3 and the coefficient of x2 is -6.

A root of a polynomial function, f(x), is a value of x for which f(x) = 0

3 2 2

1. Express each polynomial as a poduct of factors.

(a) 5 10 ( ) 9 16 ( ) 6 7 5x x b x c x x

2( ) 5 ( 2)a x x

( ) (3 4)(3 4)b x x

( ) (2 1)(3 5)c x x

22. Find the roots of the function ( ) 3 12f x x

23 12 0x 23( 4) 0x

3( 2)( 2) 0x x

2 0 2 0x x 2 2x x

Hence the roots are 2 and –2.

Division by (x – a)

Dividing a polynomial by (x – a) allows us to factorise the polynomial.

You already know how to factorise a quadratic, but how do we factorise a polynomial of degree 3 or above?

We can divide polynomials using the same method as simple division.

8 3504

323 0

3

246 350 8 43 remainder 6

Conversely, 350 8 43 6

Divisor Quotient Remainder

3 2( 2) 5 2 8 2x x x x

25x

3 25 10x x212x 8x

12x

212 24x x32x 2

32

32 64x 62

3 2 2(5 2 8 2) ( 2) ( 2)(5 12 32) 62x x x x x x x

Divisor Quotient Remainder

Note: If the remainder was zero, (x – 2) would be a factor.

This is a long winded but effective method. There is another.

3 2 2(5 2 8 2) ( 2) ( 2)(5 12 32) 62x x x x x x x

Let us look at the coefficients.

2 5 2 8 -2

5 10

12 24

32 64

62

quotientdivisor remainder

This method is called synthetic division.

If the remainder is zero, then the divisor is a factor.

Let us now look at the theory.

3 2( ) 5 2 8 2f x x x x

We divided this polynomial by (x – 2).

(2)f 62

If we let the coefficient be Q(x), then

( ) ( ) ( ) ( )f x x h Q x f h

Remainder Theorem

If a polynomial f (x) is divided by (x – h) the remainder is f (h).

We can use the remainder theorem to factorise polynomials.

Remainder Theorem

If a polynomial f (x) is divided by (x – h) the remainder is f (h).

If ( ) 0 then ( ) is a factor of ( )f h x h f x

Conversely, if ( ) is a factor of ( ) then ( ) 0.x h f x f h

4 3 21. Show that ( 4) is a factor of 2 9 5 3 4x x x x x

4 2 -9 5 -3 -4

28

-1-41

41

40

Since the remainder is zero, (x – 4) is a factor.

3 2( ) ( 4)(2 1)f x x x x x

3 22. Factorise fully 2 5 28 15x x x

To find the roots we need to consider the factors of -15.

1, 3, 5, 15

3 2 5 -28 -15

2611

335

150

2( ) ( 3)(2 11 5)f x x x x ( 3)(2 1)( 5)x x x

4 23. Factorise fully 7 18.x x

To find the roots we need to consider the factors of -18.

1, 2, 3, 6, 9, 18

3 1 0 -7 0 -18

133

92

66

3 2( ) ( 3)( 3 2 6)f x x x x x

180

-3 1 3 2 6

1-30

02

-60

2( ) ( 3)( 3)( 2)f x x x x

Finding a polynomial’s coefficientsWe can use the factor theorem to find unknown coefficients in a polynomial.

4 3 21. If ( 3) is a factor of 2 6 4 15, find . x x x px x p

Since we know (x + 3) is a factor, the remainder must be zero.

-3 2 6 p 4 -15

2-60

0p

-3p

4 - 3p

9p - 12

9p - 27 9 27 0p 9 27p

3p

4 3 22. If ( 2) and ( 4) are factors of 8, find and . x x x ax x bx a b

2 1 a -1 b -8

12

2 + a4 + 2a3 + 2a

6 + 4a

6 + 4a + b

12 + 8a + 2b

4 + 8a + 2b = 0

-4 1 a -1 b -8

1-4

a - 416 - 4a15 - 4a

16a - 60

16a + b - 60

240 - 64a - 4b

232 - 64a - 4b = 0

8 2 4

64 4 232

a b

a b

( 2) 16 4 8

64 4 232

a b

a b

48 240a

5a 40 2 4b 22b

5 and 22a b

Solving polynomial equations3 21. Find the roots of ( ) 4 6 0f x x x x

3 1 -4 1 6

13-1

-3-2

-60

3 2 24 6 ( 3)( 2)x x x x x x ( 3)( 2)( 1)x x x

Roots are 1, 2 and 3.

If we sketch the curve of f (x), we see that the roots are where f (x) crosses the X axis.

y

x

1

1

2

2

3

3

4

4

5

5

6

6

– 1

– 1

– 2

– 2

– 3

– 3

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

9

10

10

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

– 6

– 6

– 7

– 7

– 8

– 8

– 9

– 9

– 10

– 10

Functions from Graphs

f (x)

x

d

a b c

The equation of a polynomial may be established from its graph.

( ) ( )( )( )f x k x a x b x c

is found by substituting the point (0, )k d

1. From the graph, find an expression for f (x).

f (x)

x

-12

-3 2

( ) ( 3)( 2)f x k x x

Substituting (0, -12)

12 (0 3)(0 2)k 12 6k

2k ( ) 2( 3)( 2)f x x x

22( 6)x x 22 2 12x x

2. From the graph, find an expression for f (x).

f (x)

x

30

-2 1 5

( ) ( 2)( 1)( 5)f x k x x x

Substituting (0, 30)

30 (0 2)(0 1)(0 5)k 30 10k

3k

( ) 3( 2)( 1)( 5)f x x x x 23( 2)( 6 5)x x x

3 23 12 21 30x x x

3 2 23( 6 5 2 12 10)x x x x x 3 23( 4 7 10)x x x

Curve sketchingThe factor theorem can be used when sketching the graphs of polynomials.

3 21. Sketch the graph of 8 12y x x x

The Y axis intercept is (0, 12)

Y axis Intercept

We will use synthetic division to find the roots of the function. This will tell us where the graph crosses the X axis.

2 1 -1 -8 12

121

2-6

-120

2( 2)( 6)y x x x ( 2)( 2)( 3)x x x

This give us the points (0,2) and (0, -3)

X axis Intercept

Stationary Points

3 2 8 12y x x x

23 2 8dy

x xdx

0 for s.p.

(3 4)( 2) 0x x

3 4 0x 3 4x

4

3x

2 0x 2x

3 24 4 4 4

When , 8 123 3 3 3

x y

1418

27

3 2When 2, 2 2 8 2 12x y 0

Nature of Stationary Points

4 4 42 2

3 3 3x

(3 4)x ( 2)x

dy

dx

Slope

-

-

+ 0

+

-

- 0

+

+

+

4 14Maximum T.P. at ,18

3 27

Minimum T.P. at 2,0

y

x

1

1

2

2

3

3

4

4

5

5

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

2

2

4

4

6

6

8

8

10

10

12

12

14

14

16

16

18

18

20

20

– 2

– 2

– 4

– 4

4 14,18

3 27

Approximate Roots

When the roots of f (x) = 0 are not rational, we can find approximate values by an iterative process.

We know a root exists if f (x) changes sign between two values.

f (x)

xa b

f (x)

xa b

( ) 0f x at x a ( ) 0f x at x b

A root exists between a and b.

( ) 0f x at x a ( ) 0f x at x b

A root exists between a and b.

3 21. For the function ( ) 4 2 7 show that there is a real

root between 1 and 2. Find this root to two decimal places.

f x x x x

(1) 2f (2) 5f

Hence the graph crosses the x - axis between 1 and 2.

( ) Root lies betweenx f x

1 22 -5 1 and 2

1.5 -1.625 1 and 1.51.4 -0.896 1 and 1.41.3 -0.163 1 and 1.31.2 0.568 1.2 and 1.31.25 0.203 1.25 and 1.31.26 0.130 1.26 and 1.31.27 0.057 1.27 and 1.31.28 -0.016 1.27 and 1.281.275 0.020 1.275 and 1.28 Hence the root is 1.28 to 2 d.p.