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
 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 ?)
x2 + 4x – 5 = (x – 1)(x + 5)
-5 = -1  5
Factorising a quadratic
Example: Factorise 6x2 - 11x - 10.
Let f(x) = 6x2 - 11x - 10
Factors of – 10 are : a =  1,  2,  5,  10
6 + 11- 10 = 7
Let try: f(-1) = 6(-1)2 - 11 (-1) – 10 =
6 - 11- 10 = -15
Let try: f(1) = 6(1)2 - 11 (1) – 10 =
24 + 22- 10 = 36
Let try: f(-2) = 6(-2)2 - 11 (-2) – 10 =
24 - 22- 10 = -8
Let try: f(2) = 6(2)2 - 11 (2) – 10 =
150 + 55 - 10 = 195
Let try: f(-5) = 6(-5)2 - 11 (-5) – 10 =
150 - 55 - 10 = 85
Let try: f(5) = 6(5)2 - 11 (5) – 10 =
Let try: f(-10) = 6(-10)2 - 11 (-10) – 10 = 600 + 110 - 10 = 700
600 - 110 - 10 = 480
Let try: f(10) = 6(10)2 - 11 (10) – 10 =
No factors ? 6x2 does not split as x  6x
6x2 = 2x  3x
a =  ½ ,  5/2
Extended factor theorem for quadratics
If f ( ba )  0
then bx – a is a factor of f(x)
Let f(x) = 6x2 - 11x - 10
f ( 12 )  14
 2x – 1 is not a factor
f ( 12 )  3
f ( 52 )  0
 2x + 1 is not a factor
 2x - 5 is a factor
6x2 - 11x – 10 = (2x – 5)(3x ?)
6x2 - 11x – 10 = (2x – 5)(3x + 2)
- 10 = -5  2
Equating the coefficients
If ax2 + bx + c  px2 + qx + r then
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
 Step 1: Use the factor theorem to find one factor.
 Step 2: Equate the coefficients or long division.
Example:
f(2) = 0
Factorise x3 + 2x2 – 5x - 6
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 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.