GCSE Maths · Algebra

Factorising GCSE Questions and Worked Answers

Factorising writes a sum as a product. For example, 6x + 12 = 6(x + 2): both terms contain a factor of 6, and expanding the bracket returns the original expression.

Foundation & Higher6 worked examples10 original questions
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Start with the meaning

What you need to know about factorising

Suppose six bags each contain x counters and two extras. Their total can be written as 6x + 12, or as 6(x + 2). The first form counts each kind separately; the second shows six equal groups. Changing a sum into a multiplication like this is called factorising. The multiplied pieces, 6 and (x + 2), are its factors.

See the idea first

Find what each group contains

Both parts of 6x + 12 can be shared into six equal groups. Dividing each term by 6 tells us what belongs inside the bracket.

First part6x ÷ 6 = xeach group receives x
Second part12 ÷ 6 = 2each group receives 2 extras
Rebuild6(x + 2)six complete groups
From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Take out a common factor

What the problem asks: Factorise 8x + 12 fully.

How to solve it: The largest whole-number factor shared by 8 and 12 is 4. Divide both terms by 4 to get 4(2x + 3).

Take out a shared letter

What the problem asks: Factorise 5x² + 10x fully.

How to solve it: Both terms contain 5x. Dividing by 5x leaves x + 2, giving 5x(x + 2).

Make two quadratic brackets

What the problem asks: Factorise x² + 7x + 12.

How to solve it: In (x + p)(x + q), the middle coefficient is p + q and the last term is pq. Choose 3 and 4: they add to 7 and multiply to 12.

Recognise a difference of squares

What the problem asks: Factorise x² − 25.

How to solve it: 25 = 5². The product (x − 5)(x + 5) gives x² − 25 because its middle terms cancel.

A reliable routine

For x² + bx + c, after checking for a common factor

This pair-finding method applies when the coefficient of x² is 1. It follows from expanding (x + p)(x + q) = x² + (p + q)x + pq. It is not a rule for every quadratic.

  1. Find numbers whose product is c.
  2. Choose the pair whose sum is b, including their signs.
  3. Write (x + p)(x + q). A negative number makes a minus sign.
  4. Expand to verify all three terms. If no integer pair works, do not invent one.

Check: Factorising an expression is not the same as solving an equation. There is no value of x to find until a condition such as the expression equalling zero is supplied.

Fully worked

Factorising GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

A numerical common factor

2 marks
Question

Factorise 18x + 30 fully.

The highest common factor of 18 and 30 is 6.

18x+30=6(3x+5)18x+30=6(3x+5)

Nothing greater than 1 divides both remaining coefficients.

Example 2

Include the shared letter

2 marks
Question

Factorise 12x28x12x^2-8x fully.

Both terms contain 4x.

12x28x=4x(3x2)12x^2-8x=4x(3x-2)

Expanding produces 12x² − 8x. Taking out only 4 would not finish the job.

Example 3

A positive pair

2 marks
Question

Factorise x2+9x+20x^2+9x+20.

Find a pair with product 20 and sum 9: 4 and 5.

x2+9x+20=(x+4)(x+5)x^2+9x+20=(x+4)(x+5)

Check: the middle products are 5x + 4x = 9x.

Example 4

A negative product

2 marks
Question

Factorise x2+x12x^2+x-12.

The signs must differ because their product is negative. 4 × (−3) = −12 and 4 + (−3) = 1.

x2+x12=(x+4)(x3)x^2+x-12=(x+4)(x-3)

Expanding checks the +x term.

Example 5

Difference of squares

2 marks
Question

Factorise 9x2169x^2-16.

The squared quantities are 3x and 4.

9x216=(3x)2429x^2-16=(3x)^2-4^2 =(3x4)(3x+4)=(3x-4)(3x+4)

The −12x and +12x terms cancel.

Example 6

Coefficient of x² is not 1

Higher only3 marks
Question

Factorise 2x2+7x+32x^2+7x+3.

The squared term needs 2x and x; the constant needs 1 and 3. Test their placement. (2x+3)(x+1)=2x2+2x+3x+3=2x2+5x+3(2x+3)(x+1)=2x^2+2x+3x+3=2x^2+5x+3 The middle term is 5x, not 7x. Swap the constants and check all four products.

(2x+1)(x+3)=2x2+6x+x+3(2x+1)(x+3)=2x^2+6x+x+3 =2x2+7x+3=2x^2+7x+3

Therefore the factorisation is (2x + 1)(x + 3).

10 original questions · total 21 marks

Factorising GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 26 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed
1

Common number

1 mark

Factorise 9x + 15 fully.

Show worked answer

The common factor is 3.

9x+15=3(3x+5)9x+15=3(3x+5)

Divide each term by 3.

2

A subtraction

2 marks

Factorise 14y − 21 fully.

Show worked answer
14y21=7(2y3)14y-21=7(2y-3)

The minus sign stays with the second term.

3

Common letter

2 marks

Factorise 6a2+9a6a^2+9a fully.

Show worked answer

Both terms contain 3a.

6a2+9a=3a(2a+3)6a^2+9a=3a(2a+3)

Check both products.

4

Three terms

2 marks

Factorise 10x2+15x+2010x^2+15x+20 fully.

Show worked answer

All three terms contain 5, but not all contain x.

10x2+15x+20=5(2x2+3x+4)10x^2+15x+20=5(2x^2+3x+4)

Do not remove x from the constant term.

5

Positive quadratic

2 marks

Factorise x² + 8x + 15.

Show worked answer

3 × 5 = 15 and 3 + 5 = 8.

x2+8x+15=(x+3)(x+5)x^2+8x+15=(x+3)(x+5)

The pair satisfies both conditions.

6

Negative middle term

2 marks

Factorise x² − 7x + 12.

Show worked answer

The product is positive but the sum is negative, so choose −3 and −4.

x27x+12=(x3)(x4)x^2-7x+12=(x-3)(x-4)

Their sum is −7.

7

Opposite signs

2 marks

Factorise x² − 2x − 24.

Show worked answer

Choose −6 and 4: their product is −24 and sum is −2.

x22x24=(x6)(x+4)x^2-2x-24=(x-6)(x+4)
8

Squares

2 marks

Factorise 4x² − 49.

Show worked answer

The two squares are (2x)² and 7².

4x249=(2x7)(2x+7)4x^2-49=(2x-7)(2x+7)

The middle products cancel.

9

Factorise fully

Harder3 marks

Factorise 2x² + 10x + 12 fully.

Show worked answer

Take out 2 before choosing the pair.

2x2+10x+12=2(x2+5x+6)2x^2+10x+12=2(x^2+5x+6) =2(x+2)(x+3)=2(x+2)(x+3)

The bracket now has coefficient 1 on x².

10

Check cross-products

Higher only3 marks

Factorise 3x² − 10x + 3.

Show worked answer

Use first terms 3x and x. The constants must multiply to +3 and produce a negative middle term, so try −3 and −1. (3x3)(x1)=3x23x3x+3=3x26x+3(3x-3)(x-1)=3x^2-3x-3x+3=3x^2-6x+3 That gives −6x, not −10x. Swap the constants and expand again.

(3x1)(x3)=3x29xx+3(3x-1)(x-3)=3x^2-9x-x+3 =3x210x+3=3x^2-10x+3

Both the middle and final terms match.

Examiner-style feedback

Common factorising mistakes

Forgetting to divide every term

The outside factor multiplies the whole bracket. Divide each original term by that factor when constructing the inside.

Matching the product but not the sum

For x² + bx + c, the pair must satisfy both conditions. Expand to check.

Stopping too soon

A shared factor may remain inside the bracket. Check all coefficients and letter powers.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

  1. Factorising reverses expanding.
  2. Check common factors first.
  3. For a monic quadratic, match both product and sum.
  4. Expand the final factors to verify the result.
Quick answers

Factorising FAQ

Can every quadratic be factorised into integer brackets?

No. If no integer pair works, another method may be needed to solve an associated equation. Do not force incorrect factors.

Why does x² + 25 not use the difference-of-squares rule?

That rule needs subtraction. (x − 5)(x + 5) expands to x² − 25, not x² + 25.

Build connected skills

What to revise next

Content standards

Curriculum and rights review

A4: common factors, monic quadratics and difference of squares across tiers; non-monic quadratic factorisation is labelled Higher. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references