Factorising Harder Quadratics (Cambridge (CIE) IGCSE Maths)
Revision Note
Written by: Jamie Wood
Reviewed by: Dan Finlay
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Factorising Harder Quadratics
How do I factorise a quadratic expression where a ≠ 1 in ax2 + bx + c?
Method 1: Factorising by grouping
This is shown most easily through an example: factorising
We need a pair of numbers that, for
both multiply to give ac
ac in this case is 4 × -21 = -84
and both add to give b
b in this case is -25
-28 and +3 satisfy these conditions
Rewrite the middle term using -28x and +3x
Group and fully factorise the first two terms, using 4x as the common factor
and group and fully factorise the last two terms, using 3 as the common factor
These terms now have a common factor of
This whole bracket can be factorised out
This gives the answer
Method 2: Factorising using a grid
Use the same example: factorising
We need a pair of numbers that for
multiply to give ac
ac in this case is 4 × -21 = -84
and add to give b
b in this case is -25
-28 and +3 satisfy these conditions
Write the quadratic equation in a grid
(as if you had used a grid to expand the brackets)
splitting the middle term up as -28x and +3x (either order)
The grid works by multiplying the row and column headings, to give a product in the boxes in the middle
|
|
|
---|---|---|
| 4x2 | -28x |
| +3x | -21 |
Write a heading for the first row, using 4x as the highest common factor of 4x2 and -28x
|
|
|
---|---|---|
4x | 4x2 | -28x |
| +3x | -21 |
You can then use this to find the headings for the columns, e.g. “What does 4x need to be multiplied by to give 4x2?”
| x | -7 |
---|---|---|
4x | 4x2 | -28x |
| +3x | -21 |
We can then fill in the remaining row heading using the same idea, e.g. “What does x need to be multiplied by to give +3x?”
| x | -7 |
---|---|---|
4x | 4x2 | -28x |
+3 | +3x | -21 |
We can now read off the brackets from the column and row headings:
Worked Example
(a) Factorise .
We will factorise by grouping
We need two numbers that:
multiply to 6 × -3 = -18
and sum to -7
-9, and +2
Split the middle term up using these values
6x2 + 2x - 9x - 3
Factorise 2x out of the first two terms
2x(3x + 1) - 9x - 3
Factorise -3 of out the last two terms
2x(3x + 1) - 3(3x + 1)
These have a common factor of (3x + 1) which can be factorised out
(3x + 1)(2x - 3)
(b) Factorise .
We will factorise using a grid
We need two numbers that:
multiply to 10 × -7 = -70
and sum to +9
-5, and +14
Use these values to split the 9x term and write in a grid
|
|
|
---|---|---|
| 10x2 | -5x |
| +14x | -7 |
Write a heading using a common factor of 5x from the first row
|
|
|
---|---|---|
5x | 10x2 | -5x |
| +14x | -7 |
Work out the headings for the rows, e.g. “What does 5x need to be multiplied by to make 10x2?”
| 2x | -1 |
---|---|---|
5x | 10x2 | -5x |
| +14x | -7 |
Repeat for the heading for the remaining row, e.g. “What does 2x need to be multiplied by to make +14x?”
| 2x | -1 |
---|---|---|
5x | 10x2 | -5x |
+7 | +14x | -7 |
Read off the brackets from the column and row headings
(2x - 1)(5x + 7)
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