Why is "When expanding, multiplying only the first term inside the bracket, e.g. $4(2x + 3) = 8x + 3$." a common mistake in Expansion and Factorisation?

Why it happens: The eye stops after the first multiplication and the second term is left untouched. How to avoid it: The outside term must multiply every term inside: 4(2x + 3) = 8x + 12.

Why is "Forgetting the minus carries onto the second term, e.g. $-2(x + 1) = -2x + 1$." a common mistake in Expansion and Factorisation?

Why it happens: The minus sign feels like it only belongs to the first product. How to avoid it: A negative outside multiplies every term: -2(x + 1) = -2x - 2.

Why is "When multiplying two brackets, missing the middle terms, e.g. $(x + 2)(x + 3) = x^2 + 6$." a common mistake in Expansion and Factorisation?

Why it happens: Only the first and last products are noticed, and the cross terms are skipped. How to avoid it: There are four products. Use a grid: (x + 2)(x + 3) = x^2 + 5x + 6.

Why is "Taking out a factor that is not the highest, e.g. $12x + 18 = 2(6x + 9)$." a common mistake in Expansion and Factorisation?

Why it happens: Any common factor works, so a smaller one is taken before spotting the largest. How to avoid it: Find the HIGHEST common factor. 12x + 18 = 6(2x + 3), fully factorised.

Why is "Leaving a shared letter out of the common factor, e.g. $8x^2 + 12x = 4(2x^2 + 3x)$." a common mistake in Expansion and Factorisation?

Why it happens: It is easy to factorise the numbers and forget that both terms also share an x. How to avoid it: Check letters too. Both terms have an x, so 8x^2 + 12x = 4x(2x + 3).

AlgebraCambridge Lower Secondary Mathematics — Stage 8

Expansion and Factorisation

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Expansion and factorisation — Cambridge Lower Secondary Maths, Grade 7

Expanding and factorising are two sides of the same coin — one opens brackets up, the other puts them back. This guide walks you through expanding single and double brackets, multiplying two brackets together, and factorising by taking out a common factor — one short step at a time.

What you’ll learn

Mapped to the Cambridge Lower Secondary Mathematics curriculum framework.

Stage 8 — Expand expressions with a single bracket, including negative and letter terms.
Stage 8 — Expand and simplify expressions containing two or more brackets.
Stage 8 — Expand the product of two simple brackets.
Stage 8 — Factorise expressions by taking out the highest common factor.

Expanding a single bracket

Multiply every term inside the bracket by the term outside.

To expand a bracket you multiply every term inside it by the term outside. At Stage 8 the outside term can be a number, a letter, or a negative.

A negative outside the bracket changes the sign of every term.

So $-2(x - 4) = -2x + 8$ . Notice that $-2 \times -4 = +8$ — two negatives make a positive.

A few more:

$5(2y - 3) = 10y - 15$ .
$x(x + 6) = x^2 + 6x$ — a letter outside works too.
$3a(a + 2) = 3a^2 + 6a$ .

The golden rule is simple: every term inside, no exceptions. Once you can expand one bracket cleanly, the rest of this topic falls into place.

Expanding multiplies every inside term by the outside term.
A negative outside flips the sign of every term.
A letter outside works: $x(x + 6) = x^2 + 6x$ .
Two negatives multiply to a positive.

Expanding and simplifying two brackets

Expand each bracket separately, then collect the like terms.

Many expressions have two separate brackets added or subtracted, such as $3(x + 2) + 2(x + 5)$ . The method has two stages: expand each bracket, then collect like terms.

Expand both brackets, then collect like terms.

So $3(x + 2) + 2(x + 5) = 3x + 6 + 2x + 10 = 5x + 16$ .

Subtraction needs extra care. In $5(x + 3) - 2(x + 1)$ , the $-2$ multiplies both terms in the second bracket: $5x + 15 - 2x - 2 = 3x + 13.$

The most common slip is forgetting that the minus carries onto the second term — so $-2(x + 1)$ becomes $-2x - 2$ , not $-2x + 1$ . Expand fully, then collect.

Expand each bracket separately first.
Then collect the like terms across both.
A minus before a bracket multiplies every inside term.
$-2(x + 1) = -2x - 2$ .

Multiplying two brackets together

Multiply every term in the first bracket by every term in the second.

Sometimes two brackets are multiplied together, like $(x + 2)(x + 3)$ . This is different from two separate brackets — here you multiply every term in the first by every term in the second, giving four products.

A grid keeps track of all four products.

For $(x + 2)(x + 3)$ , the four products are $x^2$ , $3x$ , $2x$ and $6$ . Adding them: $x^2 + 3x + 2x + 6 = x^2 + 5x + 6.$

The two middle terms $3x$ and $2x$ are like terms, so they collect to $5x$ .

A grid (sometimes called an area model) is a tidy way to be sure you have caught all four products — one box per pairing. Multiply, fill the grid, then collect like terms.

Multiplying two brackets gives four products.
Multiply every term in the first by every term in the second.
A grid keeps track of all four products.
Collect the like middle terms at the end.

Factorising — expanding in reverse

Factorising puts a common factor back outside a bracket.

Factorising is the exact reverse of expanding. Expanding takes $3(x + 2)$ and opens it into $3x + 6$ . Factorising takes $3x + 6$ and puts it back as $3(x + 2)$ .

The same two expressions, viewed two ways.

Why bother? A factorised expression is often shorter and easier to work with — and factorising is a key skill for solving equations later.

The idea behind factorising is the common factor: a number or letter that divides into every term. In $3x + 6$ , both terms divide by 3, so 3 is a common factor. Pull it outside, and what is left goes inside the bracket.

Factorising and expanding being opposites gives you a free check: factorise, then expand your answer — if you get back where you started, you are right.

Factorising is the reverse of expanding.
It puts a common factor back outside a bracket.
A common factor divides into every term.
Check by expanding your factorised answer.

Factorising by taking out a common factor

Find the highest common factor of all terms, then divide each by it.

To factorise an expression, follow three steps:

Find the highest common factor (HCF) of every term.
Write the HCF outside a bracket.
Divide each term by the HCF — the results go inside the bracket.

Example: Factorise $6x + 9$ . The HCF of 6 and 9 is 3. So: $6x + 9 = 3(2x + 3).$ Check: $3 \times 2x = 6x$ and $3 \times 3 = 9$ . Correct.

The HCF can include a letter as well as a number.

The HCF can include a letter. For $8x^2 + 12x$ : the number HCF of 8 and 12 is 4, and both terms have an $x$ , so the HCF is $4x$ . Dividing, $8x^2 \div 4x = 2x$ and $12x \div 4x = 3$ , giving $4x(2x + 3)$ .

Always take out the highest common factor — $6x + 9 = 3(2x + 3)$ is fully factorised, but writing it as the smaller $1(6x + 9)$ has done nothing useful.

Find the highest common factor of every term.
Write the HCF outside, divide each term to fill the bracket.
The HCF can include a letter, like $4x$ .
Always take out the highest factor, not a smaller one.

Where you'll use this next

Expanding and factorising power equations, formulae and later algebra.

Getting confident with expanding and factorising now pays off across many topics:

Equations with brackets are solved by expanding first, then balancing.
Factorising becomes the main route to solving quadratic equations in later years.
Formulae are often rearranged or simplified using these exact skills.
Areas of rectangles model expansion — a rectangle $x$ by $(x + 3)$ has area $x^2 + 3x$ .

In everyday life, factorising is just spotting a shared part — like noticing that "3 tickets at $x plus a $6 fee" and "3 lots of $(x + 2)$ " describe the same total.

If expanding or factorising feels hard later, it is usually a sign slip or a missed term underneath. Come back to this guide whenever you need a refresher — there is no prize for rushing.

Equations with brackets are solved by expanding first.
Factorising leads into solving quadratic equations later.
Formulae are simplified using these skills.
Rectangle areas model the expansion of brackets.

Sources: Cambridge Lower Secondary Mathematics curriculum framework (Stage 8). Last reviewed 2026-05-19.

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Step-by-step worked examples — Expansion and Factorisation

Step-by-step solutions for expansion and factorisation, written exactly the way a tutor would explain them at the board.

1Expanding a single bracket
Getting started• expanding, brackets
Question
Expand $4(2x + 3)$ .
Step-by-step solution
1. Step 1
  Multiply the first term inside the bracket by the 4 outside.
  $4 \times 2x = 8x$
2. Step 2
  Multiply the second term inside by the 4 outside.
  $4 \times 3 = 12$
3. Step 3
  Write both results together.
  $4(2x + 3) = 8x + 12$
Answer
$8x + 12$
2Expanding with a negative outside
Getting started• expanding, brackets, negative numbers
Question
Expand $-3(x - 5)$ .
Step-by-step solution
1. Step 1
  Multiply the first term by $-3$ .
  $-3 \times x = -3x$
2. Step 2
  Multiply the second term by $-3$ . A negative times a negative is positive.
  $-3 \times (-5) = +15$
3. Step 3
  Write both results together.
  $-3(x - 5) = -3x + 15$
Answer
$-3x + 15$
3Expanding and simplifying two brackets
Building confidence• expanding, simplifying, like terms
Question
Expand and simplify $5(x + 3) - 2(x + 1)$ .
Step-by-step solution
1. Step 1
  Expand the first bracket: multiply each term by 5.
  $5(x + 3) = 5x + 15$
2. Step 2
  Expand the second bracket: the $-2$ multiplies both terms.
  $-2(x + 1) = -2x - 2$
3. Step 3
  Write both expanded brackets together.
  $5x + 15 - 2x - 2$
4. Step 4
  Collect like terms: $5x - 2x = 3x$ and $15 - 2 = 13$ .
  $3x + 13$
Answer
$3x + 13$
4Factorising by taking out a common factor
Building confidence• factorising, common factor, HCF
Question
Factorise $12x + 18$ .
Step-by-step solution
1. Step 1
  Find the highest common factor of 12 and 18. Both divide by 6.
  $\text{HCF} = 6$
2. Step 2
  Divide each term by 6 to find what goes in the bracket.
  $12x \div 6 = 2x,\quad 18 \div 6 = 3$
3. Step 3
  Write the HCF outside and the results inside a bracket.
  $12x + 18 = 6(2x + 3)$
Answer
$6(2x + 3)$
5Factorising with a letter in the common factor
Stretch• factorising, common factor, powers
Question
Factorise $10x^2 + 15x$ .
Step-by-step solution
1. Step 1
  Find the number HCF of 10 and 15. Both divide by 5.
  $\text{number HCF} = 5$
2. Step 2
  Both terms contain an $x$ , so $x$ is also a common factor. The full HCF is $5x$ .
  $\text{HCF} = 5x$
3. Step 3
  Divide each term by $5x$ .
  $10x^2 \div 5x = 2x,\quad 15x \div 5x = 3$
4. Step 4
  Write $5x$ outside the bracket. Check by expanding: $5x \times 2x = 10x^2$ and $5x \times 3 = 15x$ .
  $10x^2 + 15x = 5x(2x + 3)$
Answer
$5x(2x + 3)$

Key Definitions and Keywords — Expansion and Factorisation

The important words for expansion and factorisation and what they mean — learn these so you can explain your thinking clearly.

Expand
Key word
To multiply out a bracket by multiplying every term inside by the term outside.
Example
Expanding $3(x + 2)$ gives $3x + 6$ .
Factorise
Key word
To write an expression as a common factor multiplied by a bracket — the reverse of expanding.
Example
Factorising $6x + 9$ gives $3(2x + 3)$ .
Factor
Key word
A number or letter that divides exactly into a term with no remainder.
Common factor
Key word
A factor that divides into every term of an expression.
Example
In $4x + 8$ , the number 4 is a common factor.
Highest common factor (HCF)
Key word
The largest factor shared by all terms — the factor to take outside when factorising.
Example
The HCF of $8x^2$ and $12x$ is $4x$ .
Bracket
A pair of symbols ( ) that groups terms so they are treated as one quantity.
Term
One part of an expression, separated from the others by a $+$ or $-$ sign.
Product
The result of multiplying things together. $(x + 2)(x + 3)$ asks for a product of two brackets.
Like terms
Terms with exactly the same letter part, which can be collected together after expanding.
Coefficient
The number multiplying a letter in a term. The coefficient of $5x$ is 5.
Simplify
To write an expression in its shortest form, often by collecting like terms after expanding.
Fully factorised
An expression where the highest common factor has been taken out and nothing more can be removed.
Example
$3(2x + 3)$ is fully factorised; $1(6x + 9)$ is not.

Common Mistakes and Misconceptions — Expansion and Factorisation

The slip-ups students most often make with expansion and factorisation — and simple ways to avoid them.

✕When expanding, multiplying only the first term inside the bracket, e.g. $4(2x + 3) = 8x + 3$ .
Why it happens
The eye stops after the first multiplication and the second term is left untouched.
How to avoid it
The outside term must multiply every term inside: $4(2x + 3) = 8x + 12$ .
✕Forgetting the minus carries onto the second term, e.g. $-2(x + 1) = -2x + 1$ .
Why it happens
The minus sign feels like it only belongs to the first product.
How to avoid it
A negative outside multiplies every term: $-2(x + 1) = -2x - 2$ .
✕When multiplying two brackets, missing the middle terms, e.g. $(x + 2)(x + 3) = x^2 + 6$ .
Why it happens
Only the first and last products are noticed, and the cross terms are skipped.
How to avoid it
There are four products. Use a grid: $(x + 2)(x + 3) = x^2 + 5x + 6$ .
✕Taking out a factor that is not the highest, e.g. $12x + 18 = 2(6x + 9)$ .
Why it happens
Any common factor works, so a smaller one is taken before spotting the largest.
How to avoid it
Find the HIGHEST common factor. $12x + 18 = 6(2x + 3)$ , fully factorised.
✕Leaving a shared letter out of the common factor, e.g. $8x^2 + 12x = 4(2x^2 + 3x)$ .
Why it happens
It is easy to factorise the numbers and forget that both terms also share an $x$ .
How to avoid it
Check letters too. Both terms have an $x$ , so $8x^2 + 12x = 4x(2x + 3)$ .

Practice quiz

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Expansion and Factorisation

Detailed Notes

What you’ll learn

1Expanding a single bracket

2Expanding with a negative outside

3Expanding and simplifying two brackets

4Factorising by taking out a common factor

5Factorising with a letter in the common factor

Expand

Factorise

Factor

Common factor

Highest common factor (HCF)

Bracket

Term

Product

Like terms

Coefficient

Simplify

Fully factorised

✕When expanding, multiplying only the first term inside the bracket, e.g. 4(2x+3)=8x+34(2x + 3) = 8x + 34(2x+3)=8x+3.

✕Forgetting the minus carries onto the second term, e.g. −2(x+1)=−2x+1-2(x + 1) = -2x + 1−2(x+1)=−2x+1.

✕When multiplying two brackets, missing the middle terms, e.g. (x+2)(x+3)=x2+6(x + 2)(x + 3) = x^2 + 6(x+2)(x+3)=x2+6.

✕Taking out a factor that is not the highest, e.g. 12x+18=2(6x+9)12x + 18 = 2(6x + 9)12x+18=2(6x+9).

✕Leaving a shared letter out of the common factor, e.g. 8x2+12x=4(2x2+3x)8x^2 + 12x = 4(2x^2 + 3x)8x2+12x=4(2x2+3x).

Practice quiz

✕When expanding, multiplying only the first term inside the bracket, e.g. $4(2x + 3) = 8x + 3$ .

✕Forgetting the minus carries onto the second term, e.g. $-2(x + 1) = -2x + 1$ .

✕When multiplying two brackets, missing the middle terms, e.g. $(x + 2)(x + 3) = x^2 + 6$ .

✕Taking out a factor that is not the highest, e.g. $12x + 18 = 2(6x + 9)$ .

✕Leaving a shared letter out of the common factor, e.g. $8x^2 + 12x = 4(2x^2 + 3x)$ .