An alternative to the trial and improvement method

Factorising Difficult Quadratics

How to factorise 12x2+28x+15

Multiply the 12 and 15

Find factors of this product (180) whose sum is the coefficient of x (28).

1

2

3 12x2 + 28x + 1512x2 + 28x + 15

1

2

3

10 x 18 is 180 and

10 add 18 = 28

So our numbers are 10 and 18

18 x 10 = 180

How to factorise

Replace +28x with + 10x + 18x

Divide the expression into 2 parts

1

2

12x2 + 10x + 18x + 1512x2 + 10x + 18x + 15

We found our numbers are 10 and 18

1

2

12x2 + 10x + 18x + 15

12x2 + 28x + 15

How to factorise

Factorise the red part

Factorise the blue part

1

2

We need to factorise both parts

1

2

12x2 + 18x + 10x + 15

+6x(2x+3)

+5(2x+3)

How to factorise

One of the factors is what is in brackets

Combine what’s left for the other factor

1

2

Check that the bits inside the brackets are the same!

12

(6x + 5)

6x(2x+3)

5(2x+3)

3 Check your answer(6x + 5) (2x+3) =

12x2+28x+15

3

How to factorise 6x2+x-12

Multiply the 6 and -12

Find factors of this product (-72) whose sum is the coefficient of x (1).

1

2

3 6x2 + x – 126x2 + x – 12

1

2

3

9 x -8 is -72 and

9 add minus 8 =1

So our numbers are 9 and -8

-8 x 9 = -72

How to factorise

Replace +x with + 9x - 8x

Divide the expression into 2 parts

1

2

6x2 + 9x – 8x - 126x2 + 9x – 8x - 12

We found our numbers are +9 and - 8

1

2

6x2 + 9x – 8x - 12

6x2 + x – 12

How to factorise

Factorise the red part

Factorise the blue part

1

2

We need to factorise both parts

1

2

6x2 + 9x – 8x - 12

3x(2x+3)

-4(2x+3)

How to factorise

One of the factors is what is in brackets

Combine what’s left for the other factor

1

2

Check that the bits inside the brackets are the same!

12

(3x - 4)

3x(2x+3)

-4(2x+3)

3 Check your answer(3x - 4) (2x+3) =

6x2+x-12

3

Note

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attempts.

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get the

answer

wrong, you

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