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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