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Factorising Difficult Quadratics An alternative to the trial and improvement method How to factorise 12x2+28x+15 1 Multiply the 12 and 15 2 Find factors of this product (180) whose sum is the coefficient of x (28). 3 18 x 10 = 180 10 x 18 is 180 and 3 1 12x2 + 28x + 15 10 add 18 = 28 So our numbers are 10 and 18 2 How to factorise We found our numbers are 10 and 18 1 2 Replace +28x with + 10x + 18x 12x2 + 28x + 15 Divide the expression into 2 parts 1 12x2 + 10x + 18x + 15 12x2 + 10x + 18x + 15 2 How to factorise We need to factorise both parts 1 Factorise the red part 2 Factorise the blue part 1 +6x(2x+3) 12x2 + 18x + 10x + 15 +5(2x+3) 2 How to factorise Check that the bits inside the brackets are the same! 1 One of the factors is what is in brackets 6x(2x+3) 5(2x+3) 1 2 2 Combine what’s left for the other factor (6x + 5) 3 3 Check your answer (6x + 5) (2x+3) = 12x2+28x+15 How to factorise 6x2+x-12 1 Multiply the 6 and -12 2 Find factors of this product (-72) whose sum is the coefficient of x (1). 3 -8 x 9 = -72 9 x -8 is -72 and 3 1 6x2 + x – 12 9 add minus 8 =1 So our numbers are 9 and -8 2 How to factorise We found our numbers are +9 and - 8 1 Replace +x with + 9x - 8x 2 Divide the expression into 2 parts 6x2 + x – 12 1 6x2 + 9x – 8x - 12 6x2 + 9x – 8x - 12 2 How to factorise We need to factorise both parts 1 Factorise the red part 2 Factorise the blue part 1 3x(2x+3) 6x2 + 9x – 8x - 12 -4(2x+3) 2 How to factorise Check that the bits inside the brackets are the same! 1 One of the factors is what is in brackets 3x(2x+3) -4(2x+3) 1 2 2 Combine what’s left for the other factor (3x - 4) 3 3 Check your answer (3x - 4) (2x+3) = 6x2+x-12 Note