Factorisation of further quadratics

Sometimes we have to factorise quadratics such as 4\({x^2}\) - 19\({x}\) + 12. These are made much more difficult by the fact that there is a number (4) in front of the \({x^2}\) term (we say that the coefficient of \({x^2}\) is 4).

There are several methods you can use to solve this problem. If you already have a method that works and are comfortable with it, then continue to use that method. The following may be useful if you are looking for an alternative way to solve these types of problems.

Example

Factorise 4\({x^2}\) - 19\({x}\) + 12

Solution

First multiply 4 by 12 to obtain 48.

Secondly write down the factors of 48:

(1, 48) or (-1, -48)
(2, 24) or (-2, -24)
(3, 16) or (-3, -16)
(4, 12) or (-4, -12)
(6, 8) or (-6, -8)

Next, find the pair of factors which, when added, result in the coefficient of \({x}\), (-19).

The pair of factors we need are -3 and -16.

Now arrange the factors in a grid with the 4\({x^2}\) and 12 diagonally opposite and the -3 and -16 also diagonally opposite.

A table showing the factors in a grid with the 4x squared and 12 diagonally opposite and the -3x and -16x also diagonally opposite.

Now factorise each row and column of the grid:

A table with two columns labelled '4x' and '-3x' and two rows labelled 'x' and '-4'. The 4x squared and 12 are diagonally opposite and the -3x and -16x are also diagonally opposite.

The solution is (4\({x}\) – 3)(\({x}\) – 4).

Example

Factorise 2\({x^2}\) + 5\({x}\) + 2

Solution

First multiply 2 by 2 to obtain 4.

Secondly write down the factors of 4:

(1, 4)
(2, 2)

Next find the pair of factors which, when added, result in the coefficient of \({x}\), [5].

The pair of factors we need are 1 and 4.

Now arrange the factors in a grid with the 2\({x^2}\) and 2 diagonally opposite and the 1 and 4 also diagonally opposite.

A table showing the factors in a grid with the 2x squared and 2 diagonally opposite and the 1x and 4x also diagonally opposite.

Now factorise each row and column of the grid:

A table with two columns labelled '2x' and '1' and two rows labelled 'x' and '2'. The 2x squared and 2 are diagonally opposite and the 1x and 4x are also diagonally opposite.

Therefore, our solution is (2\({x}\) + 1)(\({x}\) + 2).

Quadratic expressions - Intermediate & Higher tier – WJECFactorisation of further quadratics