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The factor theorem. The Factor Theorem states that if f(a) = 0 for a polynomial then (x- a) is a factor of the polynomial f(x). Example. f(x) = x 2 + x - 6. f(x) = x 2 + x – 6 =. (x + 3)(x – 2) . f(-3) = (-3) 2 + (-3) – 6 =. 9 – 6 – 6 = 0. f(2) = 2 2 + 2 – 6 =. 4 + 2 – 6 = 0.

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The factor theorem

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The factor theorem l.jpg

The factor theorem

The Factor Theorem states that if f(a) = 0 for a polynomial then (x- a) is a factor of the polynomial f(x).

Example

f(x) = x2 + x - 6

f(x) = x2 + x – 6 =

(x + 3)(x – 2)

f(-3) = (-3)2 + (-3) – 6 =

9 – 6 – 6 = 0

f(2) = 22 + 2 – 6 =

4 + 2 – 6 = 0

Since f(-3) = 0 

(x + 3) is a factor of f(x) = x2 + x - 6

Since f(2) = 0 

(x - 2) is a factor of f(x) = x2 + x - 6


Factorising a quadratic l.jpg

Factorising a quadratic

 Step 1: Use the factor theorem to find one factor.

 Step 2: Use the factor to identify the second factor.

Example:

Factorise x2 + 4x - 5

Let f(x) = x2 + 4x - 5

Factors of – 5 are :

a =  1,  5

Let try: f(-1) = (-1)2 + 4 (-1) – 5 =

1 – 4 – 5 = - 8

Since f(-1)  0 

(x + 1) is not a factor of f(x)

Let try: f(1) = 12 + 4 1 – 5 =

1 + 4 – 5 = 0

Since f(1) = 0 

(x - 1) is a factor of f(x)

There is no need to go any further

x2 + 4x – 5 = (x – 1)(x ?)

-5 = -1  5

x2 + 4x – 5 = (x – 1)(x + 5)


Factorising a quadratic3 l.jpg

Factorising a quadratic

Example:

Factorise 6x2 - 11x - 10.

Let f(x) = 6x2 - 11x - 10

Factors of – 10 are :

a =  1,  2,  5,  10

Let try: f(-1) = 6(-1)2 - 11 (-1) – 10 =

6 + 11- 10 = 7

Let try: f(1) = 6(1)2 - 11 (1) – 10 =

6 - 11- 10 = -15

Let try: f(-2) = 6(-2)2 - 11 (-2) – 10 =

24 + 22- 10 = 36

Let try: f(2) = 6(2)2 - 11 (2) – 10 =

24 - 22- 10 = -8

Let try: f(-5) = 6(-5)2 - 11 (-5) – 10 =

150 + 55 - 10 = 195

Let try: f(5) = 6(5)2 - 11 (5) – 10 =

150 - 55 - 10 = 85

Let try: f(-10) = 6(-10)2 - 11 (-10) – 10 =

600 + 110 - 10 = 700

Let try: f(10) = 6(10)2 - 11 (10) – 10 =

600 - 110 - 10 = 480

No factors ?

6x2 does not split as x  6x

6x2 = 2x  3x

a =  ½ , 5/2


Extended factor theorem for quadratics l.jpg

Extended factor theorem for quadratics

then bx – a is a factor of f(x)

Let f(x) = 6x2 - 11x - 10

 2x – 1 is not a factor

 2x + 1 is not a factor

 2x - 5 is a factor

6x2 - 11x – 10 =

(2x – 5)(3x ?)

- 10 = -5  2

6x2 - 11x – 10 =

(2x – 5)(3x + 2)


Equating the coefficients l.jpg

Equating the coefficients

If ax2 + bx + c  px2 + qx + rthen

a = p, b = q and c = r

This is called equating coefficient.

Example: Given 3x + 2 is a factor of 15x2 + x – 6 . Find the other factor.

15x2 + x – 6 = (3x + 2)( )

Let the other factor to be Ax + B

15x2 + x – 6 = (3x + 2)( Ax + B)

15x2 + x – 6 = 3Ax2 + 3Bx + 2Ax + 2B

15x2 + x – 6 = 3Ax2 + (3B + 2A)x + 2B

x2: 15 = 3A 

A = 5

x: 3B + 2A = 1 

B = -3

15x2 + x – 6 = (3x + 2)(5x – 3)


Factorising a cubic l.jpg

Factorising a cubic

 Step 1: Use the factor theorem to find one factor.

 Step 2: Equate the coefficients or long division.

Example:

Factorise x3 + 2x2 – 5x - 6

f(2) =

0

By inspection

Since f(2) = 0  x - 2 is a factor of f(x).

x3 + 2x2 – 5x – 6 = (x - 2)(Ax2 + Bx + C)

= Ax3 - 2Ax2 + Bx2 - 2Bx + Cx - 2C

= Ax3 + (-2A+ B)x2 + (-2B + C)x - 2C

A =

1, -2A + B =

2, -2B + C =

-5, 2C =

-6

A =

1, B =

4, C =

3

x3 + 2x2 – 5x – 6 =(x - 2)(x2 + 4x + 3)

x3 + 2x2 – 5x – 6 =(x - 2)(x + 1)(x + 3)


Factorising a cubic7 l.jpg

Factorising a cubic

 Step 1: Use the factor theorem to find one factor.

 Step 2: Equate the coefficients or long division.

Example:

Factorise x3 + 3x2 – 12x - 14

f(1) =

- 24

f(-1) =

0

Since f(-1) = 0  x + 1 is a factor of f(x).

x3 + 3x2 – 12x – 14 = (x + 1)(Ax2 + Bx + C)

= Ax3 + Ax2 + Bx2 + Bx + Cx + C

= Ax3 + (A+ B)x2 + (B + C)x + C

A =

1, A + B =

3, B + C =

-12, C =

-14

A =

1, B =

2, C =

-14

x3 + 3x2 – 12x – 14 =(x + 1)(x2 + 2x – 14)

Quadratic does not factorize.


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