Block 2

Factors of Polynomials

What is to be learned?

• How to find factors of poLynomials of degree 3 or more.

• How we use this to solve poLynomial equations

Factors

Factors of 16? 1, 2, 4, 8, 16A divisor will be a factor if

there is no remainder

Time for The Big L x2 + 5x + 6 1 5 6– 2

1

-2

3

-6

0

D

Remainder = 0Q

(x + 3) + 0xx22 + 5x + 6 = + 5x + 6 = (x + 2)D RQ

÷ (x + 2)÷ (x + 2)

(x + 2) is a Factor

x3 – x2 + 2x – 2 1 -1 2 -2-1

1

-1

-2

2

4

-4

-6

÷ (x + 1)÷ (x + 1)

Remainder = -6(x + 1) is not a FactorIf at first you don’t succeed

try, try again…..

x3 – x2 + 2x – 2 1 -1 2 -21

1

1

0

0

2

D

Q

(x2 + 2)xx33 – x x22 + 2x + 2x – 2 2 ==(x – 1)D Q

2

0

÷ (x ÷ (x – 1) 1)

Remainder = 0(x – 1) is a Factor

May be able to factorise this bit

Factorisation of PoLynomialsusing The Big L

(also known as the remainder/factor theorem)

A divisor is a factor ifUsing Big L remainder = 0 For poLynomials of degree 3, or over.

there is no remainder

x3 + 2x2 – x – 2 1 2 -1 -21

1

1

3

3

2

D

Q

(x2 +3x + 2)xx33 + 2x + 2x22 – x – 2 – x – 2 ==(x – 1)D Q

2

0

TryTry÷ (x ÷ (x – 1) 1)

Remainder = 0(x – 1) is a Factor

factorise this bit

Solutions are factor of this

x3 + 2x2 – x – 2 1 2 -1 -21

1

1

3

3

2

D

Q

(x2 +3x + 2)xx33 + 2x + 2x22 – x – 2 – x – 2 ==(x – 1)

2

0

TryTry÷ (x - 1)÷ (x - 1)

Remainder = 0(x - 1) is a Factor

factorise this bit

Solutions are factor of this

=(x – 1)(x + 2)(x + 1)

x3 + 4x2 – 3x – 18 1 4 -3 -182

1

2

6

12

9

D

Q

(x2 +6x + 9)x3 + 4x2 – 3x – 18 == (x – 2)D Q

18

0

TryTry÷ (x ÷ (x – 2) 2)

Remainder = 0(x – 2) is a Factor

factorise this bit

Fully Factorise

= (x – 2) (x + 3)(x + 3)= (x – 2)(x + 3)2

Key Question

2x3 + px2 + 2x – 3 2 2 -3-1

2

-2

5

-5

-3

3

0

÷ (x + 1)÷ (x + 1)

Remainder = 0

If (x+ 1) is a factor, find p

p7

p = 7

What is to be learned?

• How to find factors of poLynomials of degree 3 or more.

• How we use this to solve poLynomial equations

Sammy Scientist has invented a machine that produces power (P Kw) The amount of power depends on how many hundred mice (m), he can persuade to run on the big wheel attached to it.Formula is P = 2m3 – 17m2 + 40m.How many mice will it take for the power to be 16Kw?

2m3 – 17m2 + 40m = 16 2m3 – 17m2 + 40m – 16 = 0

PoLynomial Equation

send for The Big L

2 -17 40 -16

(2m2 – 9m + 4)2m2m33 – 17m – 17m22 + 40m – 16 = 0 + 40m – 16 = 0(m – 4)

0 Remainder = 0(m – 4) is a Factor

TryTry÷ (m ÷ (m – 4) 4)

2m2m33 – 17m – 17m22 + 40m – 16 = + 40m – 16 = 00

4

2

8

-9

-36

4

16

= 0(m – 4)(2m )(m ) = 0– –1 4

m – 4 = 0 2m – 1 = 0

m – 4 = 0

m = 4or m = ½

400 or 50 mice

Solutions known as roots

The amount of power depends on how many hundred mice (m),

2 -17 40 -16

(2m2 – 9m + 4)2m2m33 – 17m – 17m22 + 40m – 16 = 0 + 40m – 16 = 0(m – 4)

0 Remainder = 0(m - 4) is a Factor

TryTry÷ (m ÷ (m – 4 4))

2m2m33 – 17m – 17m22 + 40m – 16 = + 40m – 16 = 00

4

2

8

-9

-36

4

16

= 0(m – 4)(2m )(m ) = 0– –1 4

m – 4 = 0 2m – 1 = 0

m = 4or m = ½

400 or 50 mice

Solutions known as roots

The amount of power depends on how many hundred mice (m),

2 -17 40 -16

(2m2 – 9m + 4)2m2m33 – 17m – 17m22 + 40m – 16 = 0 + 40m – 16 = 0(m – 4)

0 Remainder = 0(m - 4) is a Factor

TryTry÷ (m ÷ (m – 4 4))

2m2m33 – 17m – 17m22 + 40m – 16 = + 40m – 16 = 00

4

2

8

-9

-36

4

16

= 0(m – 4)(2m )(m ) = 0– –1 4

m – 4 = 0 2m – 1 = 0

m = 4or m = ½

400 or 50 mice

Solutions known as roots

The amount of power depends on how many hundred mice (m),

2 -17 40 -16

(2m2 – 9m + 4)2m2m33 – 17m – 17m22 + 40m – 16 = 0 + 40m – 16 = 0(m – 4)

0 Remainder = 0(m - 4) is a Factor

TryTry÷ (m ÷ (m – 4 4))

2m2m33 – 17m – 17m22 + 40m – 16 = + 40m – 16 = 00

4

2

8

-9

-36

4

16

= 0(m – 4)(2m )(m ) = 0– –1 4

m – 4 = 0 2m – 1 = 0

m = 4or m = ½

400 or 50 mice

Solutions known as roots

Solving PoLynomial Equations

• Make one side = 0• Find factor using The Big L• Fully factorise• Find solutions (roots)

(or use quadratic formula)

Ex. Show that x = 1 is a solution and solve x3 – 2x2 – x = -2

1 -2 -1 21

1

1

-1

-1

-2

(x2 – x – 2) = 0

xx33 – 2x 2x22 – x + 2 – x + 2 = 0= 0

(x – 1)

-2

0 Remainder = 0

(x – 1) is a Factor

÷ (x –– 1)

xx33 – 2x 2x22 – x + 2 = 0 – x + 2 = 0

(x – 1)(x – 2)(x + 1) = 0

x – 1 = 0, x – 2 = 0, x + 1 = 0

x = 1,x = 2, x = -1

(roots)

Ex. Show that x = -2 is a root and solve 2x3 + x2 – 5x = -2

2 1 -5 2-2

2

-4

-3

6

1

(2x2 – 3x + 1) = 0

2x2x33 + xx22 – 5x + 2 – 5x + 2 = 0= 0

(x + 2)

-2

0 Remainder = 0

(x + 2) is a Factor

÷ (x + 2+ 2)

2x3 + x2 – 5x + 2 = 0+ 2 = 0

(x + 2)(2x – 1)(x – 1) = 0

x + 2 = 0, 2x –1= 0 , x – 1 = 0

x = -2,x = ½ , x = 1

(roots)

Key Question