The factor theorem

# The factor theorem

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## The factor theorem

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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) = 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

2. 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)

3. 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

4. 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)

5. 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)

6. 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)

7. 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.