A4AlgebraFoundation & HigherGrades 3–7

Factorising

Q: Does the order of the brackets matter?

No. Multiplication can be done in either order, so (x + 4)(x - 1) = (x - 1)(x + 4).

Factorising is expanding in reverse: you put brackets back in by finding what the terms have in common. It turns a sum into a product, and products are what you need to solve quadratic equations, simplify algebraic fractions and prove results.

45 min Video by Maths Genie AQA GCSE Maths

The big picture

Every factorisation can be checked instantly by expanding it again. If $5x(x - 3)$ expands to $5x^2 - 15x$ , the factorisation is right — so there is never a reason to lose these marks.

Always look for a common factor first. It makes every later step easier: $2x^2 + 10x + 12$ becomes $2(x^2 + 5x + 6)$ , and the quadratic inside is the simple kind.

Factorising Single BracketsCurated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

What you'll learn

Factorise by taking out the highest common factor
Factorise quadratics of the form $x^2 + bx + c$
Recognise and factorise the difference of two squares
Factorise quadratics of the form $ax^2 + bx + c$ (Higher)

Common factors

Find the highest common factor (HCF) of the numbers and of the letters. Write it outside the bracket, then divide each term by it to find what goes inside.

"Factorise fully" means take out the $\textbf{highest}$ common factor. $4x^2 + 6x = 2(2x^2 + 3x)$ is not fully factorised, because $x$ is still common inside.

Example 1Factorise fully

12a + 18

1The HCF of 12 and 18 is 6.

12a \div 6 = 2a

and

18 \div 6 = 3

Answer

6(2a + 3)

Your turnSame method, new numbers — try it before moving on.

Factorise fully $20y - 35$ .

Your answer

Show worked solution

Solution

1The HCF of 20 and 35 is 5.
2 $20y \div 5 = 4y$ and $35 \div 5 = 7$ : $5(4y - 7)$ .

Answer: $5(4y - 7)$

Example 2Factorise fully

15x^2 - 10xy

1Numbers: HCF of 15 and 10 is 5. Letters: both terms contain

x

2Take out

5x

15x^2 \div 5x = 3x

and

10xy \div 5x = 2y

Answer

5x(3x - 2y)

Your turnSame method, new numbers — try it before moving on.

Factorise fully $8p^2 + 12p$ .

Your answer

Show hint

Both the numbers and the letters have something in common.

Show worked solution

Solution

1HCF of 8 and 12 is 4, and both terms contain $p$ .
2Take out $4p$ : $4p(2p + 3)$ .

Answer: $4p(2p + 3)$

Tip — Check by expanding. If the terms inside the bracket still share a factor, you have not factorised fully.

Quadratics: $x^2 + bx + c$

Look for two numbers that $\textbf{multiply}$ to give $c$ and $\textbf{add}$ to give $b$ . Those numbers go in the brackets: $x^2 + bx + c = (x + p)(x + q)$ .

Use the signs to guide you. If $c$ is positive, both numbers have the same sign as $b$ . If $c$ is negative, the numbers have different signs.

Example 1Factorise

x^2 + 7x + 12

1Pairs multiplying to 12:

1 \times 12

2 \times 6

3 \times 4

3 + 4 = 7

✓

Answer

(x + 3)(x + 4)

Your turnSame method, new numbers — try it before moving on.

Factorise $x^2 + 9x + 20$ .

Your answer

Show worked solution

Solution

1Multiply to 20 and add to 9: $4$ and $5$ .
2 $(x + 4)(x + 5)$

Answer: $(x + 4)(x + 5)$

Example 2Factorise

x^2 - 2x - 15

c = -15

is negative, so the numbers have different signs.

2Multiply to

-15

, add to

-2

-5

and

3

Answer

(x - 5)(x + 3)

Your turnSame method, new numbers — try it before moving on.

Factorise $x^2 + 3x - 28$ .

Your answer

Show hint

The product is negative, so one number is negative. They add to $+3$ .

Show worked solution

Solution

1Multiply to $-28$ and add to $3$ : $7$ and $-4$ .
2 $(x + 7)(x - 4)$

Answer: $(x + 7)(x - 4)$

Difference of two squares

An expression of the form $a^2 - b^2$ factorises as $(a + b)(a - b)$ . Expanding shows why: the middle terms $+ab$ and $-ab$ cancel.

Spot it when there are two terms, both perfect squares, with a minus between them: $x^2 - 49$ , $9y^2 - 25$ , $4 - m^2$ .

Example 1Factorise

16x^2 - 81

16x^2 = (4x)^2

and

81 = 9^2

2Use

a^2 - b^2 = (a + b)(a - b)

Answer

(4x + 9)(4x - 9)

Your turnSame method, new numbers — try it before moving on.

Factorise $x^2 - 64$ .

Your answer

Show worked solution

Solution

1 $x^2 - 8^2$ is a difference of two squares.
2 $(x + 8)(x - 8)$

Answer: $(x + 8)(x - 8)$

A calculation shortcut

The difference of two squares is a mental arithmetic trick too: $51^2 - 49^2 = (51 + 49)(51 - 49) = 100 \times 2 = 200$ . AQA sometimes sets exactly this kind of non-calculator question.

Quadratics: $ax^2 + bx + c$ (Higher)

First take out any common factor. If the $x^2$ coefficient is still not 1, multiply $a \times c$ , find two numbers that multiply to $ac$ and add to $b$ , and use them to split the middle term.

Then factorise the four terms in pairs. Both pairs must share the same bracket.

Example 1Factorise

3x^2 + 10x + 8

a \times c = 24

. Multiply to 24, add to 10:

6

and

4

2Split the middle:

3x^2 + 6x + 4x + 8

3In pairs:

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

Answer

(3x + 4)(x + 2)

Your turnSame method, new numbers — try it before moving on.

Factorise $2x^2 + 7x - 15$ .

Your answer

Show hint

$a \times c = -30$ . Find two numbers that multiply to $-30$ and add to $7$ .

Show worked solution

Solution

1 $ac = -30$ : the numbers are $10$ and $-3$ .
2 $2x^2 + 10x - 3x - 15 = 2x(x + 5) - 3(x + 5)$ .
3 $(2x - 3)(x + 5)$

Answer: $(2x - 3)(x + 5)$

Tip — $2x^2 + 10x + 12$ : take out the 2 first to get $2(x^2 + 5x + 6) = 2(x + 2)(x + 3)$ .

Think like an examiner

"Factorise fully" is a signal that there is more than one factor to take out — usually a number and a letter.
Brackets can be written in either order: $(x + 3)(x - 2)$ and $(x - 2)(x + 3)$ both score.
If a question says "factorise and solve", the factorised form earns marks on its own — write it down before solving.

Factorising patterns

ab + ac = a(b + c)

Take out the highest common factor.

x^2 + (p + q)x + pq = (x + p)(x + q)

Multiply to

c

, add to

b

a^2 - b^2 = (a + b)(a - b)

Difference of two squares.

Watch out for these

6x^2 + 9x = 3(2x^2 + 3x)

Not fully factorised —

x

is still common:

3x(2x + 3)

x^2 - 9 = (x - 3)^2

(x - 3)^2 = x^2 - 6x + 9

. The answer is

(x + 3)(x - 3)

x^2 + x - 12 = (x + 4)(x + 3)

These multiply to

+12

. You need

-12

(x + 4)(x - 3)

x^2 + 9 = (x + 3)(x - 3)

A sum of two squares does not factorise — only a difference does.

Stretch yourself

Without a calculator, work out $\dfrac{997^2 - 9}{1000}$ .

Hint — $9 = 3^2$ . Look for a difference of two squares on the top.

Reveal solution

Solution

997^2 - 3^2 = (997 + 3)(997 - 3) = 1000 \times 994

. Dividing by 1000 leaves

994

\blacksquare

Questions students ask

How do I know which method to use?

Count the terms. Two terms: common factor or difference of two squares. Three terms with

x^2

: a quadratic — take out any common factor first, then check whether the

x^2

coefficient is 1.

What if I cannot find two numbers that work?

Check the signs, and make sure you have taken out any common factor. Some quadratics, like

x^2 + x + 1

, do not factorise with whole numbers at all — then the quadratic formula is needed.

Does the order of the brackets matter?

No. Multiplication can be done in either order, so

(x + 4)(x - 1) = (x - 1)(x + 4)

Key takeaways

Take out the highest common factor of numbers and letters first.
$x^2 + bx + c$ : find two numbers that multiply to $c$ and add to $b$ .
$a^2 - b^2 = (a + b)(a - b)$ .
$ax^2 + bx + c$ : split the middle term using $ac$ , then factorise in pairs.
Always check by expanding your answer.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 33 marks

(a)Factorise $14x + 21$ .[1]
Show mark scheme (a)
- B1 $7(2x + 3)$
Answer: $7(2x + 3)$
(b)Factorise fully $9a^2b - 6ab$ .[2]
Show mark scheme (b)
- B1Partially factorised, e.g. $3(3a^2b - 2ab)$ or $ab(9a - 6)$
- B1 $3ab(3a - 2)$
Answer: $3ab(3a - 2)$

Question 2Grade 53 marks

(a)Factorise $x^2 - 11x + 24$ .[2]
Show mark scheme (a)
- M1 $(x \pm 3)(x \pm 8)$
- A1 $(x - 3)(x - 8)$
Answer: $(x - 3)(x - 8)$
(b)Hence solve $x^2 - 11x + 24 = 0$ .[1]
Show mark scheme (b)
- B1 $x = 3$ and $x = 8$ (follow through from their factors)
Answer: $x = 3$ or $x = 8$

Question 3Grade 72 marks

Factorise fully $6x^2 - 24$ .[2]
Show mark scheme
- M1Takes out the common factor: $6(x^2 - 4)$
- A1 $6(x + 2)(x - 2)$
Answer: $6(x + 2)(x - 2)$

Examiner note

$6(x^2 - 4)$ earns only one mark — "fully" means the bracket must also be factorised as a difference of two squares.

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Factorise $5x + 30$ .
- $5(x + 6)$
- $5(x + 30)$
- $x(5 + 30)$
- $6(x + 5)$
Show answer and explanation
Answer: $5(x + 6)$
HCF 5: $30 \div 5 = 6$ .
Factorise fully $6y^2 + 9y$ .
- $3y(2y + 3)$
- $3(2y^2 + 3y)$
- $y(6y + 9)$
- $6y(y + 9)$
Show answer and explanation
Answer: $3y(2y + 3)$
Take out both 3 and $y$ .
Factorise $x^2 + 8x + 15$ .
- $(x + 3)(x + 5)$
- $(x + 1)(x + 15)$
- $(x + 8)(x + 15)$
- $(x - 3)(x - 5)$
Show answer and explanation
Answer: $(x + 3)(x + 5)$
$3 \times 5 = 15$ and $3 + 5 = 8$ .
Factorise $x^2 - x - 20$ .
- $(x - 5)(x + 4)$
- $(x + 5)(x - 4)$
- $(x - 10)(x + 2)$
- $(x - 5)(x - 4)$
Show answer and explanation
Answer: $(x - 5)(x + 4)$
$-5 \times 4 = -20$ and $-5 + 4 = -1$ .
Factorise $y^2 - 100$ .
- $(y + 10)(y - 10)$
- $(y - 10)^2$
- $(y - 50)(y + 2)$
- $y(y - 100)$
Show answer and explanation
Answer: $(y + 10)(y - 10)$
Difference of two squares.
Which expression cannot be factorised?
- $x^2 + 4$
- $x^2 - 4$
- $x^2 + 4x$
- $x^2 + 4x + 4$
Show answer and explanation
Answer: $x^2 + 4$
A sum of two squares has no real factors.
Factorise $2x^2 + 5x + 3$ .
- $(2x + 3)(x + 1)$
- $(2x + 1)(x + 3)$
- $(2x + 5)(x + 3)$
- $(x + 3)(x + 1)$
Show answer and explanation
Answer: $(2x + 3)(x + 1)$
Check: $2x^2 + 2x + 3x + 3$ .
Factorise fully $3x^2 - 12x + 9$ .
- $3(x - 1)(x - 3)$
- $(3x - 3)(x - 3)$
- $3(x^2 - 4x + 3)$
- $3(x + 1)(x + 3)$
Show answer and explanation
Answer: $3(x - 1)(x - 3)$
Take out 3, then factorise $x^2 - 4x + 3$ .

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