Procedure

Factorise a quadratic by splitting the middle term

Algebra • Expressions • Factorisation

Statement

To factorise

a x^{2} + b x + c

, find two numbers whose product is

a c

and whose sum is

b

, split

b x

into those two terms, and take a common factor out of each pair; the method runs unchanged for a homogeneous quadratic

a p^{2} + b pq + c q^{2}

in two variables.

How to use it

1Compute $a c$ first, list its factor pairs, and pick the pair that sums to $b$ .
2If no integer pair works, the quadratic has no rational factorisation and the quadratic formula is the route instead.
3In two variables keep the second variable attached throughout, so the factors emerge as $(α p + β q) (γ p + δ q)$ .

Worked example

Factorise

6 x^{2} + x - 12

Step-by-step

1
$a c = 6 \times (- 12) = - 72$ , and $9 \times (- 8) = - 72$ with $9 + (- 8) = 1 = b$
Why: Find the pair whose product is $a c$ and whose sum is $b$ .
2
$6 x^{2} + 9 x - 8 x - 12 = 3 x (2 x + 3) - 4 (2 x + 3)$
Why: Split the middle term and take a common factor from each pair.
3
$= (2 x + 3) (3 x - 4)$

Answer

(2 x + 3) (3 x - 4)

Watch out

The product to match is

a c

and not

c

; forgetting the leading coefficient is the commonest failure. The two pairs must leave the same bracket behind, and if they do not, the split chosen was the wrong one.

Also known as

splitting the middle term, factorise a quadratic, factor by grouping, middle term break

Appears in past papers

Statement

Worked example

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