• Revision:Factorising

TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Factorising

In order to look at factorising, we first have a reminder of what 'expanding' means.


Expanding Brackets

Brackets should be expanded in the following ways: For an expression of the form:

\displaystyle a(b + c),

the expanded version is:

\displaystyle ab + ac,

i.e., multiply the term outside the bracket by everything inside the bracket. E.g. -

\displaystyle 2x(x + 3) = 2x^2 + 6x

[remember \displaystyle x \times x = x^2])

For an expression of the form:

\displaystyle (a + b)(c + d),

the expanded version is:

\displaystyle ac + ad + bc + bd,

in other words everything in the first bracket should be multiplied by everything in the second.


Expand \displaystyle (2x + 3)(x - 1):


\displaystyle (2x + 3)(x - 1)

\displaystyle = 2x^2 - 2x + 3x - 3

\displaystyle = 2x^2 + x - 3


Factorising is the reverse of expanding brackets, so it is putting:

\displaystyle 2x^2 + x - 3

into the form:

\displaystyle (2x + 3)(x - 1).

This is an important way of solving quadratic equations.

The first step of factorising an expression is to 'take out' any common factors which the terms have. So if you were asked to factorise x^2 + x, since x goes into both terms, you would write:

\displaystyle x(x + 1).

Factorising Quadratics

There is no simple method of factorising a quadratic expression. One way, however, is as follows:


Factorise:\displaystyle 12y^2 - 20y + 3.

\displaystyle 12y^2 - 18y - 2y + 3

[here the 20y has been split up into two numbers whose multiple is 36. 36 was chosen because this is the product of 12 and 3, the other two numbers].

The first two terms, 12y^2 and -18y both divide by 6y, so 'take out' this factor of 6y:

\displaystyle 6y(2y - 3) - 2y + 3

[we can do this because 6y(2y - 3) = 12y^2 - 18y].

Now, make the last two expressions look like the expression in the bracket:

\displaystyle 6y(2y - 3) -1(2y - 3)

The answer is: \displaystyle (2y - 3)(6y - 1)


Factorise:\displaystyle x^2 + 2x - 8.

We need to split the 2x into two numbers which multiply to give -8. This has to be 4 and -2:

\displaystyle x^2 + 4x - 2x - 8

\displaystyle = x(x + 4) - 2x - 8

\displaystyle = x(x + 4)- 2(x + 4)

\displaystyle = (x + 4)(x - 2)

Once you work out what is going on, this method makes factorising any expression easy. It is worth studying these examples further if you do not understand what is happening. Unfortunately, the only other method of factorising is by trial and error.

The Difference of Two Squares

If you are asked to factorise an expression which is one square number minus another, you can factorise it immediately. This is because:

\displaystyle a^2 - b^2 = (a + b)(a - b).


Factorise:\displaystyle 25 - x^2.

\displaystyle 25 - x^2 = (5 + x)(5 - x)

[imagine that a = 5, b = x].


This article needs other methods of factorising looking at, even if the said methods do not explain why we do what they tell us to do (an explanation can be include alongside, but need not be part of the method. Also, if we have an expanding article, then that could be linked to from here too.

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