Expanding Single and Double Brackets — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers multiplying out brackets. By the end of this guide you should be able to expand a single bracket, expand and simplify two brackets multiplied together, and handle the sign changes that arise when a bracket is subtracted.

You should also be able to expand squared brackets, recognise the difference of two squares, expand three brackets, and combine expansions with collecting like terms.

The organising idea is that expanding means every term inside meets every term outside — not some of them. A single bracket has one term outside, so each term inside is multiplied once. Two brackets have two terms in the first, so each term of the second is multiplied twice, giving four products. Three brackets give eight. Counting the products you expect before you start is the simplest check there is: if 2(x + 3) produces one term, or (x + 2)(x + 5) produces three, something has been missed. Almost every error in this topic is a product that was never formed, or a sign that was not carried.

Key terms and definitions

Expand — multiply out a bracket to remove it.

Term — a part of an expression separated by a plus or minus sign.

Coefficient — the number in front of a letter.

Like terms — terms with the same letters to the same powers, which can be collected.

Simplify — collect like terms after expanding.

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

Perfect square — a bracket multiplied by itself, such as (x + 3)².

Core concepts

Expanding a single bracket

Multiply every term inside by the term outside.

So 3(x + 4) = 3 × x + 3 × 4 = 3x + 12.

Expect as many products as there were terms inside. A bracket with two terms gives two products, so an answer with only one means a term was missed.

Signs follow the usual rules. In −2(x − 5), the outside term is negative, so both products change: −2 × x = −2x and −2 × −5 = +10, giving −2x + 10. The second sign change is the one most often dropped.

Where the term outside contains a letter, the index laws apply: x(x + 3) = x² + 3x, and 2a(3a − 4) = 6a² − 8a.

Expanding and simplifying

Many questions expand more than one bracket and then ask for the simplest form, which means collecting like terms afterwards.

For 3(x + 2) + 4(x − 1): expanding gives 3x + 6 + 4x − 4, and collecting gives 7x + 2.

Where a bracket is subtracted, the minus applies to everything inside it.

For 5(x + 3) − 2(x + 4): the second expansion is −2x − 8, not −2x + 8, so the working is 5x + 15 − 2x − 8, giving 3x + 7.

Treating the subtraction as though it only reached the first term inside is the single commonest error in the topic. Writing the −2 as the multiplier, rather than thinking of it as "subtract 2 times", makes the sign look after itself.

Expanding double brackets

Each term in the first bracket multiplies each term in the second, giving four products.

For (x + 3)(x + 5): the products are x × x = x², x × 5 = 5x, 3 × x = 3x, and 3 × 5 = 15.

That gives x² + 5x + 3x + 15, and collecting the two x terms gives x² + 8x + 15.

Any systematic method works — a grid, or the familiar FOIL order of first, outer, inner, last — provided all four products appear. The grid is the more reliable, because an empty cell is visible in a way a forgotten product is not.

Signs must travel with their terms. For (x − 4)(x + 2): the products are x², +2x, −4x and −8, giving x² − 2x − 8.

A useful check

In (x + a)(x + b), the middle coefficient is a + b and the constant is a × b.

So for (x + 3)(x + 5), the middle term should be 8x and the constant 15, which matches. For (x − 4)(x + 2), the middle is −4 + 2 = −2 and the constant is −8, which also matches.

That check takes a moment and catches sign errors immediately. It is also the pattern used in reverse when factorising, so time spent on it pays twice.

Squared brackets

A squared bracket means the bracket multiplied by itself, and it must be written out before expanding.

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

Squaring the terms separately to get x² + 9 is a serious and common error — it omits the middle term entirely. Substituting a number settles it: at x = 1, the correct expansion gives 16, while x² + 9 gives 10.

The general pattern is (x + a)² = x² + 2ax + a², so the middle coefficient is twice a. And (x − a)² = x² − 2ax + a², where the constant stays positive because a negative squared is positive.

The difference of two squares

When the two brackets are identical except for the sign between the terms, the middle terms cancel.

(x + 4)(x − 4) = x² − 4x + 4x − 16 = x² − 16.

The general pattern is (a + b)(a − b) = a² − b².

Recognising it saves time and is used constantly in factorising, in simplifying algebraic fractions, and in rationalising surd denominators.

Expanding three brackets

Expand two of them first, simplify, then multiply the result by the third.

For (x + 1)(x + 2)(x + 3): the first two give x² + 3x + 2, and multiplying that by (x + 3) means each of its three terms meets each of the two in the last bracket — six products in all.

That gives x³ + 3x² + 3x² + 9x + 2x + 6, which collects to x³ + 6x² + 11x + 6.

Working in stages, and writing the intermediate quadratic down, is what keeps this manageable. Attempting all three at once almost guarantees a lost term.

Expanding as the reverse of factorising

Every expansion can be run backwards, and seeing the two as one skill makes both easier.

Expanding turns 3(x + 4) into 3x + 12; factorising turns 3x + 12 back into 3(x + 4). Expanding turns (x + 3)(x + 5) into x² + 8x + 15; factorising reverses it.

That is why the a + b and a × b check is worth learning properly: when factorising a quadratic you are looking for two numbers with the right sum and product, which is the same relationship read from the other end.

It also gives a free check on any factorising you do — expand your answer and confirm it returns the original expression.

Expanding with two letters

The method is unchanged when more than one letter appears.

For (2x + y)(x − 3y): the products are 2x², −6xy, xy and −3y², which collects to 2x² − 5xy − 3y².

Note that xy and yx are the same term and do collect, while x² and y² do not.

Worked examples

Example 1: Subtracting a bracket

Expand and simplify 6(2x + 1) − 3(x − 4).

Expand the first bracket: 12x + 6.

Expand the second, with the minus reaching both terms: −3x + 12.

Write them together: 12x + 6 − 3x + 12.

Collect like terms: 9x + 18.

Treating the second bracket as −3x − 12 would have given 9x − 6, and the error lies entirely in that one sign.

Example 2: A squared bracket

Expand and simplify (2x − 5)².

Write it out as a product first: (2x − 5)(2x − 5).

The four products are 4x², −10x, −10x and +25.

Collecting gives 4x² − 20x + 25.

The constant is positive because −5 × −5 = +25, and the middle term is twice −10x. Squaring the terms separately would have given 4x² + 25, missing the middle term entirely.

Example 3: Three brackets

Expand and simplify (x − 2)(x + 3)(x + 1).

Take the first two: (x − 2)(x + 3) = x² + 3x − 2x − 6 = x² + x − 6.

Now multiply that by (x + 1), forming six products: x³ + x² from the x², x² + x from the x, and −6x − 6 from the constant.

So x³ + x² + x² + x − 6x − 6.

Collecting gives x³ + 2x² − 5x − 6.

Checking at x = 1: the original gives (−1)(4)(2) = −8, and the expansion gives 1 + 2 − 5 − 6 = −8. ✓

Common mistakes and how to avoid them

Multiplying only the first term inside. Every term inside meets the term outside. Count the products.

Missing the sign change when subtracting a bracket. The minus applies to everything inside.

Squaring terms separately. (x + 3)² is not x² + 9. Write the bracket out twice.

Losing a product in a double expansion. Use a grid, and expect four products.

Forgetting to collect like terms. "Expand and simplify" asks for both.

Mishandling a negative times a negative. Two negatives give a positive, which is why (x − 5)² ends with +25.

Trying three brackets in one step. Expand two, simplify, then multiply by the third.

Exam technique for "Expanding Brackets"

Count the products you expect before starting — two for a single bracket with two terms, four for a double, six or more beyond that — and check you have them all before collecting.

Use a grid for double brackets. An empty cell is visible, whereas a forgotten product is not.

Write the expansion as its own line before collecting like terms. Both stages carry marks, and an unsimplified but correct expansion still earns credit.

Check the middle coefficient and the constant against the a + b and a × b pattern for quadratics.

Substitute a simple value such as x = 1 into both the original and your answer when time allows. It catches sign errors instantly.

Read the instruction: "expand" alone does not require collecting, but "expand and simplify" does.

Quick revision summary

Every term inside meets every term outside. Count the products: two for a single bracket, four for a double, six or more for bigger expansions.

Expanding a single bracket: 3(x + 4) = 3x + 12. A negative outside changes every sign inside: −2(x − 5) = −2x + 10.

A subtracted bracket changes every sign inside it — this is the commonest error in the topic.

For double brackets, use a grid and expect four products, then collect. Check with the pattern: in (x + a)(x + b) the middle coefficient is a + b and the constant is a × b.

A squared bracket is the bracket times itself: (x + 3)² = x² + 6x + 9, not x² + 9. In general the middle coefficient is twice a.

Difference of two squares: (a + b)(a − b) = a² − b², because the middle terms cancel.

For three brackets, expand two, simplify, then multiply by the third.