1. 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 Since f(-3) = 0  (x + 3) is a factor of f(x) = x 2 + x - 6 Since f(2) = 0  (x - 2) is a factor of f(x) = x 2 + x - 6

2. Factorising a quadratic Example: Let f(x) = x 2 + 4x - 5 Factorise x 2 + 4x - 5 Since f(-1)  0  (x + 1) is not a factor of f(x)  Step 1: Use the factor theorem to find one factor.  Step 2: Use the factor to identify the second factor. 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 a factor of f(x) Let try: f(1) = 1 2 + 4  1 – 5 = 1 + 4 – 5 = 0 There is no need to go any further x 2 + 4x – 5 = (x – 1)(x ?) x 2 + 4x – 5 = (x – 1)(x + 5) -5 = -1  5

3. Factorising a quadratic Example: Let f(x) = 6x 2 - 11x - 10 Factorise 6x 2 - 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 ? 6x 2 does not split as x  6x 6x 2 = 2x  3xa =  ½,  5 / 2

4. Extended factor theorem for quadratics then bx – a is a factor of f(x) Let f(x) = 6x 2 - 11x - 10  2x – 1 is not a factor  2x + 1 is not a factor  2x - 5 is a factor 6x 2 - 11x – 10 =(2x – 5)(3x ?) - 10 = -5  2 6x 2 - 11x – 10 =(2x – 5)(3x + 2)

5. Equating the coefficients If ax 2 + bx + c  px 2 + qx + r then a = p, b = q and c = rThis is called equating coefficient. Example: Given 3x + 2 is a factor of 15x 2 + x – 6. Find the other factor. 15x 2 + x – 6 = (3x + 2)( ) Let the other factor to be Ax + B 15x 2 + x – 6 = (3x + 2)( Ax + B) 15x 2 + x – 6 = 3Ax 2 + 3Bx + 2Ax + 2B 15x 2 + x – 6 = 3Ax 2 + (3B + 2A)x + 2B x 2 : 15 = 3A  A = 5 x: 3B + 2A = 1  B = -3 15x 2 + x – 6 = (3x + 2)(5x – 3)

6. Factorising a cubic Example: Since f(2) = 0  x - 2 is a factor of f(x). Factorise x 3 + 2x 2 – 5x - 6  Step 1: Use the factor theorem to find one factor.  Step 2: Equate the coefficients or long division. A = f(2) =0By inspection x 3 + 2x 2 – 5x – 6 = (x - 2)(Ax 2 + Bx + C) = Ax 3 - 2Ax 2 + Bx 2 - 2Bx + Cx - 2C = Ax 3 + (-2A+ B)x 2 + (-2B + C)x - 2C 1, -2A + B =2, -2B + C =-5, 2C =-6 A =1, B =4, C = 3 x 3 + 2x 2 – 5x – 6 =(x - 2)(x 2 + 4x + 3) x 3 + 2x 2 – 5x – 6 =(x - 2)(x + 1)(x + 3)

7. Factorising a cubic Example: Since f(-1) = 0  x + 1 is a factor of f(x). Factorise x 3 + 3x 2 – 12x - 14  Step 1: Use the factor theorem to find one factor.  Step 2: Equate the coefficients or long division. A = f(1) =- 24f(-1) =0 x 3 + 3x 2 – 12x – 14 = (x + 1)(Ax 2 + Bx + C) = Ax 3 + Ax 2 + Bx 2 + Bx + Cx + C = Ax 3 + (A+ B)x 2 + (B + C)x + C 1, A + B =3, B + C =-12, C =-14 A =1, B =2, C =-14 x 3 + 3x 2 – 12x – 14 =(x + 1)(x 2 + 2x – 14) Quadratic does not factorize.