1MA1 · A4AlgebraFoundation & Higher

Factorising Expressions

Factorising rewrites an expression as a product. It reverses expansion and exposes algebraic structure used later to simplify expressions and solve quadratic equations.

60 min Video by Maths Genie Pearson Edexcel GCSE Maths (1MA1)

The big picture

Factorising is not a separate trick from expanding: the two processes undo one another. That gives every factorisation a built-in check—expand your brackets and confirm that the original expression returns.

Edexcel 1MA1 progresses from taking out common factors to factorising monic quadratics, difference of two squares and, at Higher tier, quadratics whose leading coefficient is not 1.

FactorisingWatch the walkthrough, then read the notes below.

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What you'll learn

Take the highest common numerical and algebraic factor out of an expression
Factorise quadratics of the form $x^{2} + b x + c$
Recognise and factorise a difference of two squares
Factorise $a x^{2} + b x + c$ at Higher tier

Taking out a common factor

Find the highest common factor of the coefficients and the lowest power of every variable shared by all terms. Put that factor outside a bracket, then divide each original term by it to fill the bracket.

A fully factorised answer uses the greatest common factor. For example, $12 x^{3} y + 18 x^{2} y^{2} = 6 x^{2} y (2 x + 3 y)$ .

ab + a c = a (b + c)

Factorising reverses the distributive law.

1The HCF of 15 and 25 is 5.

2Both terms contain at least

a^{2} b

3Divide each term by

5 a^{2} b

: the bracket contains

3 a - 5 b

Answer

5 a^{2} b (3 a - 5 b)

Why this works

Using the lowest shared power ensures the factor divides every term. Using any larger power would create negative indices inside the bracket.

Monic quadratics

To factorise $x^{2} + b x + c$ , find two numbers whose product is $c$ and whose sum is $b$ . Those numbers complete $(x + p) (x + q)$ .

Use the sign of $c$ to narrow the search. If $c$ is negative, the two numbers have opposite signs. If $c$ is positive, they have the same sign and their sum determines whether both are positive or negative.

x^{2} + b x + c = (x + p) (x + q), pq = c, p + q = b

Product

c

, sum

b

1Find factors of 12 that add to

- 7

- 3 \times - 4 = 12

and

- 3 + (- 4) = - 7

3Write one number in each bracket.

Answer

(x - 3) (x - 4)

Tip — Expand the proposed brackets mentally: the constant product and middle-term sum must both match.

Difference of two squares

A difference of two squares has two square terms separated by subtraction and no middle term. It factorises into conjugate brackets: $a^{2} - b^{2} = (a + b) (a - b)$ .

Check that both terms are perfect squares and take their square roots. For example, $25 x^{2} - 49 = (5 x + 7) (5 x - 7)$ .

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

The difference of two squares.

1Take out the common factor 3:

3 (x^{2} - 25)

2Recognise

x^{2} - 25 = x^{2} - 5^{2}

3Apply the difference-of-squares pattern.

Answer

3 (x + 5) (x - 5)

Why this works

When the conjugate brackets expand, the middle terms $+ ab$ and $- ab$ cancel, leaving exactly $a^{2} - b^{2}$ .

Non-monic quadratics (Higher)

For $a x^{2} + b x + c$ , multiply $a$ and $c$ . Find two numbers that multiply to $a c$ and add to $b$ , split the middle term using those numbers, then factorise by grouping.

a \times c = 6 \times 3 = 18

; find two numbers with product 18 and sum 11: 9 and 2.

2Split the middle term:

6 x^{2} + 9 x + 2 x + 3

3Factorise pairs:

3 x (2 x + 3) + 1 (2 x + 3)

4Take out the repeated bracket.

Answer

(3 x + 1) (2 x + 3)

Tip — If all three original terms share a factor, take it out before using the split-middle-term method.

Think like an examiner

“Factorise fully” means continue until no common factor or recognised factorisation remains.
For a quadratic, your brackets must reproduce all three terms when expanded. A quick expansion is the most reliable check.
In Higher-tier non-monic questions, show the split middle term and grouping; these stages can earn method marks.

Remember these

ab + a c = a (b + c)

Common-factor pattern.

x^{2} + b x + c = (x + p) (x + q)

Monic quadratic, where

p + q = b

and

pq = c

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

Difference of two squares.

Watch out for these

Stopping at

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

when a larger common factor exists.

Always check whether the expression inside the bracket can be factorised further.

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

This is a difference of squares:

(x + 3) (x - 3)

Choosing numbers that multiply to

c

but do not add to

b

Both conditions must hold; expand the brackets to verify.

Stretch yourself

Factorise fully $12 x^{3} - 27 x$ .

Hint — There is a common factor first, and the expression left inside is a difference of two squares.

Questions students ask

How do I know whether an expression is fully factorised?

Check that no factor remains common to every term inside any bracket and that no bracket can be factorised using a quadratic or difference-of-squares pattern.

Why do I multiply

a

and

c

for a non-monic quadratic?

Splitting the middle term into values with product

a c

creates two pairs with a matching binomial factor, allowing factorisation by grouping.

Key takeaways

Take out the greatest common numerical and algebraic factor first.
For $x^{2} + b x + c$ , find two numbers with product $c$ and sum $b$ .
Recognise $a^{2} - b^{2}$ and use conjugate brackets.
For $a x^{2} + b x + c$ , split the middle term using factors of $a c$ , then group.

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