Factorising Quadratic Expressions — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers writing a quadratic expression as a product of brackets. By the end of this guide you should be able to factorise x² + bx + c by finding two numbers, handle every combination of signs, and recognise the difference of two squares.

You should also be able to factorise when the x² coefficient is greater than 1, take out a common factor first where one exists, spot a perfect square, and know when an expression will not factorise at all.

The organising idea is that factorising is expanding run backwards, so the check is built into the method. Expanding (x + 3)(x + 5) gives x² + 8x + 15, where the middle coefficient is the two numbers added and the constant is the two numbers multiplied. Factorising simply asks the question from the other end: which two numbers add to 8 and multiply to 15? Every technique here is that relationship, and every answer can be verified in seconds by expanding it back. Nobody need ever be uncertain whether a factorisation is right.

Key terms and definitions

Factorise — write an expression as a product of factors, usually brackets.

Expand — the reverse: multiply the brackets out.

Quadratic expression — one whose highest power of x is 2, written ax² + bx + c.

Coefficient — the number in front of a term.

Factor pair — two numbers multiplying to give a particular product.

Difference of two squares — the pattern a² − b², which factorises to (a + b)(a − b).

Perfect square — a quadratic factorising into two identical brackets.

Common factor — something dividing into every term.

Core concepts

Always look for a common factor first

Before anything else, check whether every term shares a factor, and take it outside.

For 2x² + 10x + 12, every term divides by 2, giving 2(x² + 5x + 6). The bracket then factorises further to 2(x + 2)(x + 3).

Missing this step makes the remaining work harder than it needs to be, and an answer left as a partial factorisation rarely earns full marks.

Letters count as common factors too: 3x² + 12x becomes 3x(x + 4).

Factorising x² + bx + c

When the x² coefficient is 1, find two numbers that multiply to c and add to b, then write them in the brackets.

For x² + 7x + 12: two numbers multiplying to 12 and adding to 7 are 3 and 4, so the answer is (x + 3)(x + 4).

Listing the factor pairs of c systematically is more reliable than guessing. For 12 the pairs are 1 and 12, 2 and 6, 3 and 4 — and only the last pair adds to 7.

The order of the brackets does not matter; (x + 4)(x + 3) is the same answer.

Working out the signs

The two signs follow from b and c, and thinking it through beats memorising cases.

If c is positive, the two numbers multiply to a positive, so they share a sign — and that sign is the sign of b. So x² + 7x + 12 gives (x + 3)(x + 4), while x² − 7x + 12 gives (x − 3)(x − 4).

If c is negative, the two numbers multiply to a negative, so they have opposite signs, and the larger of the two takes the sign of b. So x² + 2x − 15 gives (x + 5)(x − 3), while x² − 2x − 15 gives (x − 5)(x + 3).

That last distinction — which number takes the minus — is where most sign errors happen, and expanding the answer back settles it immediately.

Factorising ax² + bx + c

When the x² coefficient is greater than 1, the two numbers must multiply to a × c rather than to c alone, while still adding to b. Then split the middle term and factorise by grouping.

For 2x² + 7x + 3: here a × c = 6, and the numbers multiplying to 6 and adding to 7 are 6 and 1.

Split the middle term: 2x² + 6x + x + 3.

Group into pairs and factorise each: 2x(x + 3) + 1(x + 3).

The bracket (x + 3) is common to both, so take it out: (2x + 1)(x + 3).

Two details matter. The order of the split does not affect the answer — writing 2x² + x + 6x + 3 leads to the same brackets by a different route. And the two brackets produced by the grouping must be identical; if they are not, either the factor pair or the grouping is wrong.

Watch the sign when the second group starts with a minus. For 3x² − 10x + 8, splitting gives 3x² − 6x − 4x + 8, and the second group factorises as −4(x − 2), not +4(x − 2) — taking out the negative is what makes the brackets match.

Difference of two squares

When a quadratic has no x term and is a subtraction of two squares, it factorises into a matching pair:

a² − b² = (a + b)(a − b)

So x² − 25 = (x + 5)(x − 5), and x² − 49 = (x + 7)(x − 7).

It appears in disguised forms. 4x² − 9 = (2x + 3)(2x − 3), since 4x² is (2x)² and 9 is 3². And 9x² − 16y² = (3x + 4y)(3x − 4y).

The pattern requires a subtraction. There is no equivalent factorisation for x² + 25, and questions include such expressions precisely to see whether you know that.

Sometimes a common factor reveals the pattern: 2x² − 18 becomes 2(x² − 9), which then factorises to 2(x + 3)(x − 3).

Perfect squares

Where the two numbers turn out to be the same, the expression factorises into two identical brackets and can be written as a square.

x² + 6x + 9 = (x + 3)(x + 3) = (x + 3)².

The recognisable pattern is that c is a perfect square and b is twice its root. In x² + 6x + 9, the constant 9 is 3² and the middle coefficient 6 is 2 × 3.

Likewise x² − 10x + 25 = (x − 5)², since 25 is 5² and 10 is 2 × 5.

Spotting these saves time and connects directly to completing the square.

When a quadratic will not factorise

Factorising only works when the roots are rational, and many quadratics simply have no whole-number factor pair.

For x² + 3x + 5, the pairs multiplying to 5 are 1 and 5 — which add to 6, not 3. No other pair exists, so the expression does not factorise.

Rather than searching indefinitely, list the factor pairs of c, check each sum, and if none works, stop. Where the question is to solve an equation, switch to the quadratic formula or completing the square.

Checking by expanding

Every factorisation can be verified in one line by multiplying the brackets back out.

For (x + 5)(x − 3): the four products are x², −3x, +5x and −15, which collect to x² + 2x − 15. If that matches the original, the factorisation is right.

This check is quick, complete and catches every sign error, which is why uncertainty in this topic is avoidable.

Worked examples

Example 1: A negative constant

Factorise x² − 3x − 40.

The constant is negative, so the two numbers have opposite signs. They must multiply to −40 and add to −3.

The factor pairs of 40 are 1 and 40, 2 and 20, 4 and 10, 5 and 8. The pair differing by 3 is 5 and 8.

Since the sum is negative, the larger number takes the minus: −8 and +5.

So the answer is (x − 8)(x + 5).

Check by expanding: x² + 5x − 8x − 40 = x² − 3x − 40. ✓

Example 2: A common factor and a difference of two squares

Factorise 3x² − 27 fully.

Every term divides by 3, so take it out first: 3(x² − 9).

The bracket is a difference of two squares, since 9 is 3²: x² − 9 = (x + 3)(x − 3).

So the full factorisation is 3(x + 3)(x − 3).

The word "fully" is the signal that more than one stage is needed. Stopping at 3(x² − 9) would lose a mark.

Example 3: A coefficient in front of x²

Factorise 6x² + 11x + 3.

Here a × c = 6 × 3 = 18, so find two numbers multiplying to 18 and adding to 11. Those are 9 and 2.

Split the middle term: 6x² + 9x + 2x + 3.

Group and factorise each pair: 3x(2x + 3) + 1(2x + 3).

The common bracket is (2x + 3), so the answer is (2x + 3)(3x + 1).

Check by expanding: 6x² + 2x + 9x + 3 = 6x² + 11x + 3. ✓

Common mistakes and how to avoid them

Missing a common factor. Check every term for a shared factor before starting.

Getting the signs the wrong way round. A negative constant means opposite signs, with the larger number taking the sign of b.

Using c instead of a × c when the x² coefficient is not 1. The product target changes; the sum target does not.

Mismatched brackets after grouping. The two brackets must be identical — usually a sign has been mishandled in the second group.

Trying to factorise a sum of two squares. x² + 25 does not factorise; the pattern needs a subtraction.

Leaving a factorisation incomplete. "Factorise fully" means every stage.

Searching indefinitely. List the factor pairs, check the sums, and stop if none works.

Exam technique for "Factorising Quadratic Expressions"

Check for a common factor as your first written step, every time.

Write out the factor pairs of the target product and their sums, so the search is visible and systematic rather than guesswork.

State the two numbers you have found before writing the brackets, since that identification carries the method mark.

Expand your answer to check it. It takes one line and removes all doubt.

Look for the words "fully" and "completely" — they indicate more than one stage of factorising.

If the expression is a difference of two squares or a perfect square, use that pattern directly rather than searching for a factor pair.

Quick revision summary

Factorising is expanding run backwards, so every answer can be checked by multiplying out.

Take out a common factor first, including letters: 2x² + 10x + 12 = 2(x + 2)(x + 3).

For x² + bx + c, find two numbers that multiply to c and add to b. If c is positive the two share the sign of b; if c is negative they have opposite signs, with the larger taking the sign of b.

For ax² + bx + c, find two numbers multiplying to a × c and adding to b, split the middle term, and factorise by grouping — the two brackets produced must match.

Difference of two squares: a² − b² = (a + b)(a − b), so 4x² − 9 = (2x + 3)(2x − 3). It needs a subtraction; x² + 25 does not factorise.

A perfect square has c as a square and b as twice its root: x² + 6x + 9 = (x + 3)².

If no factor pair works, the expression does not factorise — use the quadratic formula or completing the square instead.