In this section we consider examples of the process of factorising, whereby the process of removing brackets is reversed and brackets are introduced into expressions.

Factorise:

(a)

8x + 12

Note that both terms are multiples of 4, so we can write,

8x + 12 = 4(2x + 3)

(b)

35x + 28

Here both terms are multiples of 7, so

35x + 28 = 7(5x + 4)

Results like this can be checked by multiplying out the bracket to get back to the original expression.

Factorise,

(a)

x^2 + 2x

Here, as both terms are multiples of *x*, we can write,

x^2 + 2x = x(x + 2)

(b)

3x^2 - 9x

In this case, both terms are multiples of *x* and 3, giving,

3x^2 - 9x = 3x(x - 3)

(c)

x^3 - x^2

In this example, both terms are multiples of *x*^{2} ,

x^3 - x^2 = x^2 (x - 1)

Sometimes it is possible to factorise in stages. For example, in part (b), you could have worked like this:

3x^2 - 9x | = 3(x^2 - 3x) |

= 3x(x - 3) |

Factorise:

(a)

x^2 + 9x + 18

This expression will need to be factorised into two brackets:

x^2 + 9x + 18 = (x )(x )

As the expression begins *x*^{2}, both brackets must begin with *x*. The two numbers to go in the brackets must multiply together to give 18 and add to give 9. So they must be 3 and 6, giving,

x^2 + 9x + 18 = (x + 3)(x + 6)

You can check this result by multiplying out the brackets.

(b)

x^2 + 2x - 15

We note first that two brackets are needed and that both must contain an *x*, as shown:

x^2 + 2x - 15 = (x )(x )

Two other numbers are needed which, when multiplied give –15 and when added give 2. In this case, these are –3 and 5. So the factorisation is,

x^2 + 2x - 15 = (x - 3)(x + 5)

Check this result by multiplying out the brackets.

(c)

x^2 - 7x + 12

Again, we begin by noting that,

x^2 - 7x + 12 = (x )(x )

We require two numbers which, when multiplied give 12 and when added give –7. In this case, these numbers are –3 and –4.

x^2 - 7x + 12 = (x - 3)(x - 4)

Note: To type indeces on this page use ^ sign. e.g. *n*^{2}: