1.3PureFoundation

Factorising Expressions

Factorising is expanding in reverse: you rewrite an expression as a product of brackets. It is the key that unlocks solving quadratics, simplifying fractions and sketching curves, so it is worth getting fast and accurate here.

35 min Video by Zeeshan Zamurred Algebraic Expressions

Edexcel AS Level Maths: 1.3 FactorisingWatch the full walkthrough before the notes below.

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What you'll be able to do

Take out the highest common factor of an expression
Factorise quadratics of the form x² + bx + c
Factorise quadratics where the coefficient of x² is greater than 1
Recognise and factorise the difference of two squares
Choose the right factorising method for a given expression

Taking out a common factor

Always check for a common factor first. Look for the largest number and the highest power of each variable that divides into every term, then place it outside a bracket.

You can check your answer instantly by expanding — it should return the original expression.

ab + a c = a (b + c)

The shared factor

a

comes out the front.

1The common factor of

6

and

9

3

; both terms contain

x

2So the factor is

3 x

3Divide each term by

3 x

6 x^{2} \div 3 x = 2 x

and

9 x \div 3 x = 3

Answer

3 x (2 x + 3)

Tip — Take out the HIGHEST common factor in one go. Pulling out just part of it (e.g. 3 instead of 3x) means more work and lost marks.

Factorising x² + bx + c

When the coefficient of $x^{2}$ is $1$ , you need two numbers that $multiply$ to give $c$ (the constant) and $add$ to give $b$ (the coefficient of $x$ ).

Pay attention to signs: a positive $c$ means both numbers share a sign; a negative $c$ means they have opposite signs.

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

Find two numbers that add to

b

and multiply to

c

1Need two numbers multiplying to

12

and adding to

7

2Those are

3

and

4

Answer

(x + 3) (x + 4)

Tip — List the factor pairs of c, then see which pair adds to b. It turns guesswork into a quick checklist.

When the x² coefficient is bigger than 1

For $a x^{2} + b x + c$ with $a > 1$ , use the $split-the-middle$ method: find two numbers that multiply to $a \times c$ and add to $b$ , split the middle term, then factorise in pairs.

1Multiply

a \times c = 2 \times 3 = 6

. Need two numbers multiplying to

6

, adding to

7

: that is

6

and

1

2Split the middle term:

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

3Factorise in pairs:

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

4The bracket

(x + 3)

is common, so factor it out.

Answer

(2 x + 1) (x + 3)

Tip — If your two pairs do not share an identical bracket, recheck the numbers — one sign is usually wrong.

Difference of two squares

Any expression of the form “square minus square” factorises into a conjugate pair. Spotting this pattern is much faster than the general method.

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

Two squares subtracted ⟶ a pair of brackets with opposite signs.

1Recognise

25 = 5^{2}

, so this is

x^{2} - 5^{2}

Answer

(x + 5) (x - 5)

Tip — Difference of two squares only works for subtraction. a² + b² does not factorise with real numbers.

Formula recap

ab + a c = a (b + c)

Common factor first — always.

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

p + q = b and pq = c.

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

Difference of two squares.

2 x^{2} + 7 x + 3 = (2 x + 1) (x + 3)

Harder quadratics: split the middle term.

Common mistakes to avoid

Taking out only part of the common factor: 6x² + 9x = 3(2x² + 3x).

Take out the full HCF 3x: 6x² + 9x = 3x(2x + 3).

Getting the signs wrong, e.g. x² − 2x − 15 = (x − 3)(x + 5).

The numbers must add to −2 and multiply to −15: (x − 5)(x + 3).

Trying to factorise a sum of two squares such as x² + 9.

x² + 9 does not factorise over the real numbers — leave it.

Key takeaways

Check for a common factor before anything else.
For x² + bx + c: two numbers that multiply to c and add to b.
For ax² + bx + c (a > 1): split the middle term and factorise in pairs.
Difference of two squares a² − b² = (a + b)(a − b) — subtraction only.

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