It is also possible to factorise expressions such as
x2 + 5x + 6
to obtain
(x + 2)(x + 3)
First consider what happens when two brackets are multiplied together. For example,
(x + 2)(x + 3) = x2 + 5x + 6.
Note that the 5 is given by 2 + 3 and the 6 is given by 2 × 3.
When factorising a quadratic like this we need to find two numbers which, when added together, give one number and when multiplied together give the other number.
For example, when factorising
x2 + 8x + 12
we need two numbers which give 12 when multiplied and 8 when added. These are, of course, 2 and 6. Hence x2 + 8x + 12 = (x + 2)(x + 6).
Worked Examples
Factorise x2 + 9x + 20.
The solution will be of the form
(x + a)(x + b)
where a × b = 20 and a + b = 9.
You may immediately see that the two numbers are 4 and 5. However, it will not always be obvious. A helpful approach is to write the possible pairs of numbers which multiply to give 20.
| x2 + 9x + 20 = (x + | )(x + | ) | ||
| 1 | 20 | |||
| 2 | 10 | |||
| 4 | 5 |
It is then easy to see that only the third pair of numbers add up to 9. So
x2 + 9x + 20 = (x + 4)(x + 5)
Factorise x2 − 3x − 10.
The solution will be of the form (x )(x ). Considering the ways of obtaining −10 by multiplication gives
| x2 − 3x − 10 = (x | )(x | ) | ||
| −1 | +10 | |||
| −2 | +5 | |||
| +1 | −10 | |||
| +2 | −5 |
Only the fourth possibility gives a total of −3 when the two terms are added, so
x2 − 3x − 10 = (x + 2)(x − 5)
Factorise x2 − 5x + 6.
The solution will be of the form (x )(x ). Considering ways of obtaining +6 (including negative factors since the x component has a negative coefficient) gives:
| x2 − 5x + 6 = (x | )(x | ) | ||
| +6 | +1 | |||
| +3 | +2 | |||
| −6 | −1 | |||
| −3 | −2 |
The last of these gives a total of −5 when the two terms are added, so
x2 − 5x + 6 = (x − 3)(x − 2)
Factorise 2x2 − x − 3.
The solution will be of the form (2x )(x ) to give the 2x2 term. Considering the ways of obtaining −3 gives:
| 2x2 − x − 3 = (2x | )(x | ) | |||
| −1 | +3 | Note that these will be multiplied by the 2 in the 2x term | |||
| +1 | −3 | ||||
| +3 | −1 | ||||
| −3 | +1 |
From the last of these we can obtain the middle term,
| (−3 × x) + (1 × 2x) | = −3x + 2x | |
| = −x | ||
| so | ||
| 2x2 − x − 3 | = (2x − 3)(x + 1) |
You can check that these brackets multiply out to give the original expression.
Factorise completely
x2 − xy
x2 − xy = x(x − y)
e2 − 1
e2 − 1 = (e − 1)(e + 1)
5p2 + 9pq − 2q2
5p2 + 9pq − 2q2 = (5p − q)(p + 2q)
(The other possibilities which do not give the correct RHS are
(5p + q) (p − 2q) = 5p2 − 9pq − 2q2
(5p + 2q) (p − q) = 5p2 − 3pq − 2q2
(5p − 2q) (p + q) = 5p2 + 3pq − 2q2 )
Note
It is often a good idea to check the answers you obtain by expanding the brackets.
You may remember that
This result is known as the difference between two squares and can be used to factorise some expressions.
Factorise the following using the difference between two squares result.
x2 − 9
| x2 − 9 | = x2 − 32 |
| = (x + 3)(x − 3) |
4x2 − 25
| 4x2 − 25 | = (2x)2 − 52 |
| = (2x − 5)(2x + 5) |
Exercises
