What is the difference between expansion and factorising in algebra?

In algebra, expansion refers to the process of multiplying out brackets to express an expression as a sum or difference of terms. For example, expanding (x + 2)(x - 3) results in x^2 - x - 6. Factorising is the reverse process, where you express a polynomial as a product of its factors. For instance, factorising x^2 - x - 6 gives (x + 2)(x - 3). Both processes are fundamental in simplifying expressions and solving equations in the Cambridge Lower Secondary Mathematics curriculum.

How do you expand algebraic expressions with multiple brackets?

To expand algebraic expressions with multiple brackets, apply the distributive property, also known as the FOIL method for binomials. For example, to expand (x + 3)(x - 2), multiply each term in the first bracket by each term in the second bracket: x*x, x*(-2), 3*x, and 3*(-2). This results in x^2 - 2x + 3x - 6, which simplifies to x^2 + x - 6. This method is essential for mastering algebraic manipulation in the Cambridge Lower Secondary Mathematics curriculum.

What are the common techniques for factorising quadratic expressions?

Common techniques for factorising quadratic expressions include finding two numbers that multiply to the constant term and add to the coefficient of the linear term. For example, to factorise x^2 + 5x + 6, find two numbers that multiply to 6 and add to 5, which are 2 and 3. Thus, the expression can be factorised as (x + 2)(x + 3). Other methods include completing the square and using the quadratic formula, which are all part of the Cambridge Lower Secondary Mathematics curriculum.

Why is factorising important in solving algebraic equations?

Factorising is crucial in solving algebraic equations because it allows you to simplify expressions and find the roots of equations. By expressing a quadratic equation in its factorised form, such as (x + 2)(x - 3) = 0, you can easily determine the solutions by setting each factor to zero, resulting in x = -2 and x = 3. This method is a key component of algebraic problem-solving in the Cambridge Lower Secondary Mathematics curriculum.

Can you explain how to factorise expressions with common factors?

To factorise expressions with common factors, identify the greatest common factor (GCF) of all terms in the expression and factor it out. For example, in the expression 4x^2 + 8x, the GCF is 4x. Factor it out to get 4x(x + 2). This technique simplifies expressions and is a foundational skill in the Cambridge Lower Secondary Mathematics curriculum, helping students to efficiently solve algebraic problems.

Why is "Multiplying only the first inside term when expanding, e.g. writing $4(x + 3) = 4x + 3$." a common mistake in Expansion and Factorising?

Why it happens: Attention drops once the first multiplication is done. How to avoid it: Picture an arrow from the outside factor to each inside term in turn — every term gets multiplied.

Why is "Forgetting the middle terms when expanding double brackets, e.g. writing $(x + 3)(x + 5) = x^2 + 15$." a common mistake in Expansion and Factorising?

Why it happens: Only the First and Last products of FOIL are remembered; Outside and Inside go missing. How to avoid it: Always write all four products before tidying: x^2 + 5x + 3x + 15 = x^2 + 8x + 15.

Why is "Forgetting the sign flip when a negative is outside the bracket, e.g. $-2(y - 5) = -2y - 10$." a common mistake in Expansion and Factorising?

Why it happens: The negative is only applied to the first inside term. How to avoid it: A negative outside flips every inside sign: -2(y - 5) = -2y + 10.

Why is "Not taking out the biggest common factor, e.g. factorising $4x^2 + 6x$ as $2(2x^2 + 3x)$." a common mistake in Expansion and Factorising?

Why it happens: The number factor is taken out but the shared x is missed. How to avoid it: Check the inside bracket for further common factors. Here x still divides both terms, so the full factorisation is 2x(2x + 3).

Why is "Getting the signs wrong in a quadratic factorisation, e.g. factorising $x^2 + 2x - 15$ as $(x - 5)(x + 3)$." a common mistake in Expansion and Factorising?

Why it happens: The two numbers are chosen but their signs are swapped. How to avoid it: If the constant is negative, the two numbers have opposite signs; the larger one matches the sign of the middle term. Here +5 and -3 give a sum of +2, so the answer is (x + 5)(x - 3).

AlgebraCambridge Lower Secondary Mathematics — Checkpoint

Expansion and Factorising

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Expansion and Factorising — Cambridge Lower Secondary Maths, Checkpoint

Revise how to expand single and double brackets, and how to factorise expressions by spotting common factors and simple quadratic patterns — building confidence with both directions so your algebra stays tidy heading into your Checkpoint.

What you’ll learn

Mapped to the Cambridge Lower Secondary Mathematics curriculum framework.

Expand single brackets and double brackets reliably, including with negatives.
Collect like terms after expanding to give a tidy answer.
Factorise expressions by taking out a common factor.
Factorise simple quadratic expressions of the form $x^2 + bx + c$ .

Expanding a single bracket

Multiply every term inside the bracket by the factor outside — none gets left behind.

Expanding a bracket means rewriting it without the brackets, by multiplying every term inside by the factor outside. The rule that powers this is the distributive law: $a(b + c) = ab + ac$ .

A few quick examples set the pattern:

$4(x + 3) = 4x + 12$
$-3(2y - 5) = -6y + 15$ (the $-3$ multiplies both inside terms)
$x(x + 7) = x^2 + 7x$ (a letter factor uses the index law)

Every inside term is multiplied — none stays untouched.

After expanding, collect like terms if the bigger expression has any. For example $4(x + 3) + 2(x - 5) = 4x + 12 + 2x - 10 = 6x + 2$ . Two warnings worth remembering: a minus outside a bracket flips every inside sign, and a missing operation between the factor and the bracket is always multiplication.

Multiply every inside term by the outside factor.
A negative outside factor flips every inside sign.
A letter factor uses the index law: $x \times x = x^2$ .
Expand first, then collect like terms.

Expanding double brackets (FOIL)

Multiply every term in the first bracket by every term in the second — four products in total.

When two brackets sit side by side, like $(x + 3)(x + 5)$ , you have to multiply every term in the first by every term in the second. The friendly memory trick is FOIL:

First terms: $x \times x = x^2$
Outside terms: $x \times 5 = 5x$
Inside terms: $3 \times x = 3x$
Last terms: $3 \times 5 = 15$

Add the four products together: $x^2 + 5x + 3x + 15 = x^2 + 8x + 15$ . The two middle terms always collect into one.

Four products from the four pairings — then add the like terms.

Watch the signs carefully. For $(x - 4)(x + 6)$ , the four products are $x^2$ , $+6x$ , $-4x$ and $-24$ , giving $x^2 + 2x - 24$ . The "Last" pair is the constant term and inherits the signs of both bracketed numbers.

Use FOIL — First, Outside, Inside, Last.
Get four products from every double-bracket expansion.
Add the Outside and Inside products into one middle term.
Keep every sign attached to its number throughout.

Factorising by a common factor

Take out the biggest factor that divides every term — number and letters.

Factorising is the reverse of expanding. Instead of removing brackets, you put them back by spotting what each term has in common.

For an expression like $6x + 9$ , both terms share a factor of $3$ . Take it out: $6x + 9 = 3(2x + 3)$ . The inside of the bracket is what is left after dividing each original term by $3$ .

When letters are involved too, take the largest number that divides every coefficient and the lowest power of each letter that appears in every term.

$4x^2 + 6x = 2x(2x + 3)$ — common factor $2x$ .
$10y^3 - 15y^2 = 5y^2(2y - 3)$ — common factor $5y^2$ .

Take out the biggest factor — number and letters — that divides every term.

After factorising, check by expanding your bracket back out. If it returns to the original expression, you've factorised completely. If you can still see another common factor inside the bracket, you have not gone far enough.

Factorising is the reverse of expanding.
Take the largest number that divides every coefficient.
Take the lowest power of each letter that appears in every term.
Check by expanding your answer back out.

Factorising simple quadratics

$x^2 + bx + c$ factorises to $(x + p)(x + q)$ , where $p + q = b$ and $pq = c$ .

A simple quadratic is an expression of the form $x^2 + bx + c$ — three terms, with $x^2$ as the highest power and no number in front of the $x^2$ . These factorise into a pair of brackets $(x + p)(x + q)$ , where two clever numbers $p$ and $q$ must satisfy:

$p \times q = c$ (their product is the constant), and
$p + q = b$ (their sum is the coefficient of $x$ ).

So to factorise $x^2 + 7x + 12$ , look for two numbers that multiply to $12$ and add to $7$ . Try the pairs: $1$ and $12$ (no, adds to $13$ ), $2$ and $6$ (no, adds to $8$ ), $3$ and $4$ (yes, adds to $7$ ). So $x^2 + 7x + 12 = (x + 3)(x + 4)$ .

List pairs, work out their sums, and pick the matching row.

Signs matter. If $c$ is positive, both $p$ and $q$ have the same sign as $b$ . If $c$ is negative, $p$ and $q$ have opposite signs, and the bigger one carries the sign of $b$ .

$x^2 - 5x + 6 = (x - 2)(x - 3)$ — both negative because $c$ positive, $b$ negative.
$x^2 + 2x - 15 = (x + 5)(x - 3)$ — opposite signs because $c$ negative; $+5$ wins because $b$ is positive.

As always, check by expanding your factorised answer.

$x^2 + bx + c$ factorises to $(x + p)(x + q)$ .
Find $p$ and $q$ : their product is $c$ and their sum is $b$ .
If $c > 0$ , both signs match the sign of $b$ .
If $c < 0$ , the signs are opposite; the bigger number takes the sign of $b$ .

Checking and tidying your answer

Expand your factorised answer back out — it should match what you started with.

Whether you've expanded or factorised, the single most useful check is to do the opposite operation and see whether you land back where you began.

If you expanded $(x + 3)(x + 5)$ and got $x^2 + 8x + 15$ , factorise it back: two numbers that multiply to $15$ and add to $8$ — that's $3$ and $5$ , giving $(x + 3)(x + 5)$ . The loop closes.

If you factorised $4x^2 + 6x$ and got $2x(2x + 3)$ , expand: $2x \times 2x = 4x^2$ , $2x \times 3 = 6x$ . Returns to the original — the answer is good.

One last habit. After factorising, scan the inside of the bracket for any further common factor. If you can spot one, factorise again. The aim is fully factorised, not just partly so.

Always check by doing the inverse operation.
If expanding, factorise back to check.
If factorising, expand back to check.
Look for any further common factor in the inside bracket.

Where you'll use this next

Expanding and factorising are the engines of all later algebra.

Confident expansion and factorising open the door to:

Solving quadratic equations — most start with a factorised form, then use the rule "if $AB = 0$ then $A = 0$ or $B = 0$ ".
Algebraic fractions — common factors cancel from top and bottom only if both are factorised first.
Identities and proofs — proving two expressions are equal often comes down to expanding both sides.
Simplifying formulae — many physics, chemistry and finance formulae need factorising before rearranging.

If a later topic feels slow, it is often a missed sign in an expansion or an incomplete factorisation underneath. Come back to this guide whenever you need a steady reset — accuracy beats speed every time.

Quadratic equations start with a factorised form.
Algebraic fractions need factors on top and bottom.
Identities are proved by expanding both sides.
Many real-world formulae factorise before they rearrange.

Sources: Cambridge Lower Secondary Mathematics curriculum framework. Last reviewed 2026-05-19.

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Step-by-step worked examples — Expansion and Factorising

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

1Expanding a single bracket
Getting started• expand, single bracket
Question
Expand $4(x + 3)$ .
Step-by-step solution
1. Step 1
  Multiply each term inside the bracket by the outside factor.
  $4 \times x = 4x,\quad 4 \times 3 = 12$
2. Step 2
  Write the two products as one expression.
  $4x + 12$
Answer
$4x + 12$
2Factorising by a common factor
Getting started• factorise, common factor
Question
Factorise $6x + 9$ completely.
Step-by-step solution
1. Step 1
  Find the largest number that divides both $6$ and $9$ . It is $3$ .
2. Step 2
  Take $3$ outside a bracket and divide each original term by $3$ to fill the bracket.
  $6x \div 3 = 2x,\quad 9 \div 3 = 3$
3. Step 3
  Write the factorised form and check by expanding.
  $3(2x + 3)$
Answer
$3(2x + 3)$
3Expanding two brackets with FOIL
Building confidence• expand, double bracket, foil
Question
Expand and simplify $(x + 3)(x + 5)$ .
Step-by-step solution
1. Step 1
  First terms.
  $x \times x = x^2$
2. Step 2
  Outside and Inside terms.
  $x \times 5 = 5x,\quad 3 \times x = 3x$
3. Step 3
  Last terms.
  $3 \times 5 = 15$
4. Step 4
  Collect the two middle terms.
  $5x + 3x = 8x$
5. Step 5
  Write the tidy answer.
  $x^2 + 8x + 15$
Answer
$x^2 + 8x + 15$
4Factorising with a variable in the common factor
Building confidence• factorise, common factor, letters
Question
Factorise $4x^2 + 6x$ completely.
Step-by-step solution
1. Step 1
  Find the largest number that divides both coefficients: $2$ .
2. Step 2
  Both terms contain at least one $x$ , so the lowest power of $x$ in common is $x$ .
3. Step 3
  The common factor is $2x$ . Divide each term by $2x$ .
  $4x^2 \div 2x = 2x,\quad 6x \div 2x = 3$
4. Step 4
  Write the factorised form.
  $4x^2 + 6x = 2x(2x + 3)$
Answer
$2x(2x + 3)$
5Factorising a simple quadratic
Stretch• factorise, quadratic
Question
Factorise $x^2 + 2x - 15$ .
Step-by-step solution
1. Step 1
  Look for two numbers whose product is $-15$ and whose sum is $+2$ .
2. Step 2
  Try pairs that multiply to $-15$ : $1 \times -15$ , $-1 \times 15$ , $3 \times -5$ , $-3 \times 5$ .
3. Step 3
  Check which pair sums to $+2$ .
  $-3 + 5 = 2 \checkmark$
4. Step 4
  Write the factorised form using those two numbers.
  $x^2 + 2x - 15 = (x + 5)(x - 3)$
5. Step 5
  Check by expanding: $x^2 - 3x + 5x - 15 = x^2 + 2x - 15$ ✓
Answer
$(x + 5)(x - 3)$

Key Definitions and Keywords — Expansion and Factorising

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

Expand
Key word
To remove brackets by multiplying each inside term by the outside factor. So $3(x + 4) = 3x + 12$ .
Factorise
Key word
To rewrite an expression as a product of factors — the reverse of expanding. So $6x + 9 = 3(2x + 3)$ .
Common factor
Key word
A factor (number or letter) that divides exactly into every term of an expression. The largest one is taken out when factorising.
Bracket
The round symbols $($ and $)$ that group terms together. Anything multiplying a bracket applies to every term inside it.
FOIL
Key word
A memory aid for expanding double brackets: First, Outside, Inside, Last — the four pairings of terms whose products are added together.
Quadratic expression
Key word
An expression whose highest power is $x^2$ , for example $x^2 + 7x + 12$ .
Coefficient
The number in front of a variable. In $7x$ the coefficient is $7$ .
Constant term
A term in an expression that has no variable — just a number. In $x^2 + 7x + 12$ the constant term is $12$ .
Distributive law
The rule that lets you expand a bracket: $a(b + c) = ab + ac$ . Multiplication distributes over addition.
Like terms
Terms with the same variable parts and the same powers. After expanding, only like terms can be combined.
Product
The result of multiplying terms together. In factorising, you write an expression as a product of factors.

Common Mistakes and Misconceptions — Expansion and Factorising

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

✕Multiplying only the first inside term when expanding, e.g. writing $4(x + 3) = 4x + 3$ .
Why it happens
Attention drops once the first multiplication is done.
How to avoid it
Picture an arrow from the outside factor to each inside term in turn — every term gets multiplied.
✕Forgetting the middle terms when expanding double brackets, e.g. writing $(x + 3)(x + 5) = x^2 + 15$ .
Why it happens
Only the First and Last products of FOIL are remembered; Outside and Inside go missing.
How to avoid it
Always write all four products before tidying: $x^2 + 5x + 3x + 15 = x^2 + 8x + 15$ .
✕Forgetting the sign flip when a negative is outside the bracket, e.g. $-2(y - 5) = -2y - 10$ .
Why it happens
The negative is only applied to the first inside term.
How to avoid it
A negative outside flips every inside sign: $-2(y - 5) = -2y + 10$ .
✕Not taking out the biggest common factor, e.g. factorising $4x^2 + 6x$ as $2(2x^2 + 3x)$ .
Why it happens
The number factor is taken out but the shared $x$ is missed.
How to avoid it
Check the inside bracket for further common factors. Here $x$ still divides both terms, so the full factorisation is $2x(2x + 3)$ .
✕Getting the signs wrong in a quadratic factorisation, e.g. factorising $x^2 + 2x - 15$ as $(x - 5)(x + 3)$ .
Why it happens
The two numbers are chosen but their signs are swapped.
How to avoid it
If the constant is negative, the two numbers have opposite signs; the larger one matches the sign of the middle term. Here $+5$ and $-3$ give a sum of $+2$ , so the answer is $(x + 5)(x - 3)$ .

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Expansion and Factorising

Short Study Notes in the form of Slides

Short Study Notes — Expansion and Factorising

Detailed Notes

What you’ll learn

1Expanding a single bracket

2Factorising by a common factor

3Expanding two brackets with FOIL

4Factorising with a variable in the common factor

5Factorising a simple quadratic

Expand

Factorise

Common factor

Bracket

FOIL

Quadratic expression

Coefficient

Constant term

Distributive law

Like terms

Product

✕Multiplying only the first inside term when expanding, e.g. writing 4(x+3)=4x+34(x + 3) = 4x + 34(x+3)=4x+3.

✕Forgetting the middle terms when expanding double brackets, e.g. writing (x+3)(x+5)=x2+15(x + 3)(x + 5) = x^2 + 15(x+3)(x+5)=x2+15.

✕Forgetting the sign flip when a negative is outside the bracket, e.g. −2(y−5)=−2y−10-2(y - 5) = -2y - 10−2(y−5)=−2y−10.

✕Not taking out the biggest common factor, e.g. factorising 4x2+6x4x^2 + 6x4x2+6x as 2(2x2+3x)2(2x^2 + 3x)2(2x2+3x).

✕Getting the signs wrong in a quadratic factorisation, e.g. factorising x2+2x−15x^2 + 2x - 15x2+2x−15 as (x−5)(x+3)(x - 5)(x + 3)(x−5)(x+3).

Practice questions

Practice quiz

Assess your understanding

Expansion and Factorising — frequently asked questions

✕Multiplying only the first inside term when expanding, e.g. writing $4(x + 3) = 4x + 3$ .

✕Forgetting the middle terms when expanding double brackets, e.g. writing $(x + 3)(x + 5) = x^2 + 15$ .

✕Forgetting the sign flip when a negative is outside the bracket, e.g. $-2(y - 5) = -2y - 10$ .

✕Not taking out the biggest common factor, e.g. factorising $4x^2 + 6x$ as $2(2x^2 + 3x)$ .

✕Getting the signs wrong in a quadratic factorisation, e.g. factorising $x^2 + 2x - 15$ as $(x - 5)(x + 3)$ .