What is factorising in the context of Edexcel IGCSE Mathematics?

Factorising is the process of breaking down an expression into a product of simpler expressions, or factors, that when multiplied together give the original expression. In the Edexcel IGCSE Mathematics curriculum, factorising is an essential skill used to simplify equations and solve quadratic equations.

How do you factorise a quadratic expression in Edexcel IGCSE Mathematics?

To factorise a quadratic expression of the form ax^2 + bx + c, you need to find two numbers that multiply to ac and add to b. Once these numbers are identified, you can rewrite the middle term and factor by grouping. This method is crucial for solving quadratic equations in the Edexcel IGCSE syllabus.

What is the difference between factorising and expanding in Mathematics?

Factorising involves breaking down an expression into its factors, while expanding is the process of multiplying factors to form an expression. In the Edexcel IGCSE Mathematics curriculum, students learn both skills to manipulate and solve equations effectively.

Why is factorising important in solving equations in Edexcel IGCSE Mathematics?

Factorising is important because it simplifies complex equations, making them easier to solve. For instance, factorising a quadratic equation allows you to set each factor to zero and solve for the variable. This technique is a fundamental part of the Equations, Formulae and Identities topic in the Edexcel IGCSE Mathematics course.

Can you provide an example of factorising a simple algebraic expression?

Certainly! Consider the expression x^2 - 5x + 6. To factorise, find two numbers that multiply to 6 and add to -5. These numbers are -2 and -3. Rewrite the expression as (x - 2)(x - 3). This factorised form is useful for solving equations and is a key skill in the Edexcel IGCSE Mathematics curriculum.

Why is "Stopping after one round of factorising" a common mistake in Factorising?

Why it happens: Student factors out a common factor and forgets to check for more. How to avoid it: After EACH step, ask: 'can this factor further?' Difference of squares? Quadratic pattern? Common factor? Stop only when ALL pieces are irreducible. Source: 4MA1/1H — examiner reports 2022-2024

Why is "Trying to factorise $x^{2} + 9$ as $(x + 3)(x + 3)$" a common mistake in Factorising?

Why it happens: Confusing sum of squares with difference. How to avoid it: a^2 + b^2 DOESN'T factor over the reals. Only a^2 - b^2 = (a-b)(a+b). Test: (x+3)(x+3) = x^2 + 6x + 9 x^2 + 9.

Why is "Pulling out only PART of the common factor" a common mistake in Factorising?

Why it happens: Easier to spot small factors. How to avoid it: Find the LARGEST common factor: HCF of all coefficients × HCF of all variable parts. 6x^2 + 9x HCF = 3x, not 3 or x.

Why is "Wrong signs in quadratic factorising" a common mistake in Factorising?

Why it happens: Not testing the result. How to avoid it: After factoring, MULTIPLY OUT to verify. (x - 2)(x + 7)? Expand: x^2 + 7x - 2x - 14 = x^2 + 5x - 14. ✓

Equations, Formulae and IdentitiesEdexcel IGCSE Mathematics (4MA1)

Factorising

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Factorising Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

Reverse of expanding. Common factor → difference of squares → quadratic. Master the order and 'fully factorise' becomes routine.

What you’ll learn

Mapped to the Cambridge IGCSE 4MA1 syllabus (2026 onwards).

2.3 — Factorise expressions by extracting common factors.
2.3 — Factorise expressions of the form $a^{2} - b^{2}$ .
2.3 — Factorise quadratic expressions in $x^{2} + bx + c$ and $ax^{2} + bx + c$ forms.
2.3 — Factorise fully — apply multiple techniques.

Common factor — always start here

Find HCF of all terms. Pull out. Always check first.

The HCF method:

Find the HCF of all coefficients (numbers).
Find the HCF of all variable parts (use lowest power).
Multiply: that's your common factor.
Divide each term by it; what remains stays in the bracket.

Example. Factorise $6x^{2} + 9x$ .

HCF of $6, 9$ : $3$ .
HCF of $x^{2}, x$ : $x$ .
Common factor: $3x$ .
Pull out: $3x(2x + 3)$ .

Multi-variable example. Factorise $12x^{3}y - 8x^{2}y^{2}$ .

HCF of $12, 8$ : $4$ .
HCF of $x^{3}, x^{2}$ : $x^{2}$ .
HCF of $y, y^{2}$ : $y$ .
Common factor: $4x^{2}y$ .
Pull out: $4x^{2}y(3x - 2y)$ .

With negative leading term. Factorise $-2x^{2} + 8x$ .

Pull out $-2x$ : $-2x(x - 4)$ .
(Always pull out a NEGATIVE if the leading term is negative.)

Worked qualitative. Why pull out the LARGEST common factor?

$6x^{2} + 9x$ could be written as $3(2x^{2} + 3x)$ — but the inner bracket can be factored further.
Better: $3x(2x + 3)$ — fully factored in one step.
Edexcel awards full marks only for the fully factored form.

Edexcel tip. ALWAYS check for a common factor BEFORE applying other techniques. Even if it's just a sign change, pulling out makes the rest easier.

HCF of numbers × HCF of variables.
Largest common factor.
Pull negative if leading term is negative.
Always do this FIRST.

Difference of two squares

$a^{2} - b^{2} = (a-b)(a+b)$ . Pattern is unmistakable.

The pattern: TWO terms, both perfect squares, with a MINUS in between.

$\boxed{a^{2} - b^{2} = (a - b)(a + b)}$

Examples:

$x^{2} - 16 = (x - 4)(x + 4)$ .
$9 - y^{2} = (3 - y)(3 + y)$ .
$25x^{2} - 4 = (5x - 2)(5x + 2)$ .
$a^{2} - b^{2}c^{2} = (a - bc)(a + bc)$ .

Worked qualitative. Why doesn't $a^{2} + b^{2}$ factor?

Try: $(a + b)^{2} = a^{2} + 2ab + b^{2}$ , has middle term.
Try: $(a + bi)(a - bi) = a^{2} + b^{2}$ — uses imaginary $i = \sqrt{-1}$ , beyond IGCSE.
Over the reals, $a^{2} + b^{2}$ is irreducible.

Combined factor. Often need common factor first.

Example: $4x^{2} - 36$ .

Common factor: $4(x^{2} - 9)$ .
Difference of squares: $(x - 3)(x + 3)$ .
Combined: $4(x - 3)(x + 3)$ .

Algebraic difference of squares.

$x^{4} - y^{4}$ is a difference of squares with $a = x^{2}, b = y^{2}$ :

$x^{4} - y^{4} = (x^{2} - y^{2})(x^{2} + y^{2})$ .
$x^{2} - y^{2}$ factors further: $(x - y)(x + y)$ .
Final: $(x - y)(x + y)(x^{2} + y^{2})$ .

Edexcel tip. When you see a $-$ between two squares, factor immediately. Mark schemes credit recognising the pattern.

$a^{2} - b^{2} = (a-b)(a+b)$ .
Both terms must be perfect squares.
Minus sign between them.
Apply repeatedly if needed.

Factorising quadratics with $a = 1$

Find pair multiplying to $c$ , adding to $b$ .

Form: $x^{2} + bx + c$ .

Method: Find two numbers $p, q$ such that:

$p \times q = c$ .
$p + q = b$ .

Then $x^{2} + bx + c = (x + p)(x + q)$ .

Example. $x^{2} + 5x + 6$ .

Need $p \times q = 6$ and $p + q = 5$ .
Pairs: $(1, 6), (2, 3)$ . Sum $5$ ? Yes — $(2, 3)$ .
Factor: $(x + 2)(x + 3)$ .

Sign cases.

$b$	$c$	$p, q$ signs
+	+	both positive
-	+	both negative
+	-	one positive, one negative; positive larger
-	-	one positive, one negative; negative larger

Examples:

$x^{2} + 5x - 14$ :

$p \times q = -14$ . $p + q = +5$ .
$(7, -2)$ . Sum $= +5$ . ✓
$(x + 7)(x - 2)$ .

$x^{2} - 8x + 12$ :

$p \times q = +12$ . $p + q = -8$ .
Both negative. $(-2, -6)$ . Sum $-8$ . ✓
$(x - 2)(x - 6)$ .

$x^{2} - 5x - 14$ :

$p \times q = -14$ . $p + q = -5$ .
$(-7, 2)$ . Sum $-5$ . ✓
$(x - 7)(x + 2)$ .

Worked qualitative. What if no integer factor pair works?

The quadratic doesn't factor (over integers).
Use the quadratic formula instead.
Or completing the square.

Edexcel tip. List factor pairs systematically — write all of them and circle the right pair. Avoid 'guess and check'.

Multiply to $c$ , add to $b$ .
Sign of $c$ tells signs of $p, q$ .
$+c$ : same sign. $-c$ : opposite signs.
If no factor: use formula.

Factorising quadratics with $a > 1$ — the AC method

Find pair multiplying to $ac$ , adding to $b$ . Split middle, factor by grouping.

Form: $ax^{2} + bx + c$ where $a > 1$ .

Method (AC method):

Compute $a \times c$ .
Find two numbers $p, q$ with $p \times q = ac$ and $p + q = b$ .
Split middle term: $ax^{2} + px + qx + c$ .
Factor by grouping (pull out common factor from each pair).

Example. Factorise $2x^{2} + 7x + 3$ .

$ac = 2 \times 3 = 6$ . $b = 7$ .
Pair multiplying to $6$ , adding to $7$ : $(1, 6)$ .
Split: $2x^{2} + x + 6x + 3$ .
Group: $x(2x + 1) + 3(2x + 1)$ .
Factor: $(2x + 1)(x + 3)$ .

Example with negative. Factorise $3x^{2} - 11x + 6$ .

$ac = 18$ . $b = -11$ . Both factors negative.
Pair: $(-9, -2)$ . Sum $-11$ . ✓
Split: $3x^{2} - 9x - 2x + 6$ .
Group: $3x(x - 3) - 2(x - 3)$ .
Factor: $(x - 3)(3x - 2)$ .

Test by expanding.

$(x - 3)(3x - 2) = 3x^{2} - 2x - 9x + 6 = 3x^{2} - 11x + 6$ . ✓

When does the AC method work? When the quadratic factors over integers. Otherwise, use the formula.

Worked qualitative. Why does grouping work?

After splitting: $2x^{2} + x + 6x + 3$ .
Pair 1: $2x^{2} + x = x(2x + 1)$ .
Pair 2: $6x + 3 = 3(2x + 1)$ .
Both have $(2x + 1)$ as a common factor — that's the bridge.
Factor: $(2x + 1)(x + 3)$ .

The AC method ENSURES this common factor will appear.

Edexcel tip. Show every step of the AC method. Mark schemes typically award M1 for splitting, M1 for grouping, A1 for the answer.

AC method: $p \times q = ac$ , $p + q = b$ .
Split middle, factor by grouping.
Test by expanding back.
If no integer factor: use the formula.

Fully factorise — applying multiple techniques

Common factor → difference of squares → quadratic. Apply in order.

'Fully factorise' = until no more factoring is possible.

Standard order:

Pull out the largest common factor.
Check for difference of two squares.
Check for quadratic factoring.
Check the remaining pieces — recurse.

Example. Fully factorise $4x^{2} - 36$ .

Common factor: $4(x^{2} - 9)$ .
Difference of squares: $4(x - 3)(x + 3)$ .
No more factors.
Final: $4(x - 3)(x + 3)$ .

Example. Fully factorise $2x^{3} - 18x$ .

Common factor: $2x(x^{2} - 9)$ .
Difference of squares: $2x(x - 3)(x + 3)$ .
Final: $2x(x - 3)(x + 3)$ .

Example. Fully factorise $x^{4} - 16$ .

Difference of squares: $(x^{2} - 4)(x^{2} + 4)$ .
$x^{2} - 4$ factors again: $(x - 2)(x + 2)$ .
$x^{2} + 4$ doesn't factor over reals.
Final: $(x - 2)(x + 2)(x^{2} + 4)$ .

Worked qualitative. A student factors $9x^{2} - 36$ as $9(x^{2} - 4) = 9(x - 2)(x + 2)$ . Another factors as $(3x - 6)(3x + 6)$ . Are both correct?

Check: $(3x - 6)(3x + 6) = 9x^{2} - 36$ . ✓
But $(3x - 6) = 3(x - 2)$ and $(3x + 6) = 3(x + 2)$ , so this equals $9(x - 2)(x + 2)$ . Same thing.
Either is correct. But the first is cleaner — Edexcel would credit it more.

Edexcel tip. When a question says 'factorise FULLY' and you write a partial factoring, you lose the A1. Always check if pieces can factor further.

Common factor → DOTS → quadratic.
Recurse on remaining pieces.
Test by expanding to verify.
Fully factor = no more reducible.

How it’s examined

Factorising appears every Paper 1H AND 2H (3-6 marks). Often a stepping stone to solving equations or simplifying. Examiner reports flag (1) not factoring fully, (2) sign errors in quadratic factoring, (3) trying $a^{2} + b^{2}$ as a difference of squares.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

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Step-by-step worked examples — Factorising

Step-by-step solutions to past-paper-style questions on factorising, written exactly the way a tutor would explain them at the board.

1Factorise out a common factor
Foundation• common factor
Question
Factorise $6x^{2} + 9x$ .
Step-by-step solution
1. Step 1
  Find the HCF of the terms. Numbers: HCF( $6, 9$ ) $= 3$ . Variables: HCF( $x^{2}, x$ ) $= x$ .
2. Step 2
  HCF: $3x$ . Pull it out:
  $6x^{2} + 9x = 3x(2x + 3)$
Answer
$3x(2x + 3)$
Examiner tip
Always pull out the LARGEST common factor. Mark schemes give M1 for common factor identified, A1 for fully factored. Doing $3(2x^{2} + 3x)$ wouldn't earn the A1.
2Difference of two squares
Higher• Adapted from 4MA1/1H Jan 2024 Q9• difference of squares
Question
Factorise $9x^{2} - 25$ .
Step-by-step solution
1. Step 1
  Recognise pattern: $a^{2} - b^{2} = (a-b)(a+b)$ .
2. Step 2
  Identify $a^{2} = 9x^{2}$ → $a = 3x$ . $b^{2} = 25$ → $b = 5$ .
3. Step 3
  Apply:
  $9x^{2} - 25 = (3x - 5)(3x + 5)$
Answer
$(3x - 5)(3x + 5)$
Examiner tip
Difference of two squares = $(a^{2} - b^{2}) = (a-b)(a+b)$ . Always identify $a$ and $b$ explicitly.
3Factorise a quadratic with leading coefficient 1
Higher• quadratic, factorising
Question
Factorise $x^{2} + 5x - 14$ .
Step-by-step solution
1. Step 1
  Find two numbers that MULTIPLY to $-14$ and ADD to $+5$ .
2. Step 2
  Factor pairs of $-14$ : $(1, -14)$ , $(-1, 14)$ , $(2, -7)$ , $(-2, 7)$ . Sums: $-13, 13, -5, 5$ . The pair $(-2, 7)$ gives $5$ .
3. Step 3
  Factorise:
  $x^{2} + 5x - 14 = (x - 2)(x + 7)$
Answer
$(x - 2)(x + 7)$
Examiner tip
For $x^{2} + bx + c$ : find two numbers that multiply to $c$ , add to $b$ . List factor pairs systematically.
4Factorise a quadratic with leading coefficient > 1
Higher• Adapted from 4MA1/2H May/Jun 2024 Q12• quadratic, non-monic
Question
Factorise $2x^{2} + 7x + 3$ .
Step-by-step solution
1. Step 1
  Find two numbers that MULTIPLY to $2 \times 3 = 6$ and ADD to $+7$ .
2. Step 2
  Factor pairs of $6$ : $(1, 6)$ , $(2, 3)$ . The pair $(1, 6)$ gives $7$ .
3. Step 3
  Split the middle term using these:
  $2x^{2} + 7x + 3 = 2x^{2} + x + 6x + 3$
4. Step 4
  Factor by grouping:
  $x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$
Answer
$(2x + 1)(x + 3)$
Examiner tip
For $ax^{2} + bx + c$ when $a > 1$ : AC method. Find two numbers multiplying to $ac$ , adding to $b$ . Split the middle, factor by grouping.
5Factorise fully — common factor + difference of squares
Higher• multi-step factorising
Question
Factorise fully $4x^{2} - 36$ .
Step-by-step solution
1. Step 1
  Pull out the common factor first: $4x^{2} - 36 = 4(x^{2} - 9)$ .
2. Step 2
  $x^{2} - 9$ is difference of squares: $(x - 3)(x + 3)$ .
3. Step 3
  Combine:
  $4x^{2} - 36 = 4(x - 3)(x + 3)$
Answer
$4(x - 3)(x + 3)$
Examiner tip
ALWAYS pull out common factors first. 'Factorise FULLY' means apply all techniques in turn until no more.

Key Formulae — Factorising

The formulae you need to memorise for factorising on the Pearson Edexcel IGCSE 4MA1 paper, with every variable defined in plain English and a note on when to use it.

Difference of two squares
$a^{2} - b^{2} = (a - b)(a + b)$
$a, b$
any expressions
When to use
When you see two perfect squares with a subtraction. The classic factor pattern.
Example
$x^{2} - 16 = (x - 4)(x + 4)$ . $9y^{2} - 4 = (3y - 2)(3y + 2)$ .

Key Definitions and Keywords — Factorising

Definitions to memorise and the exact keywords mark schemes credit for factorising answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4MA1 sitting.

Factor (algebra)
Examiner keyword
An expression that divides another expression exactly. Found by 'reverse multiplication'.
Example
$3x$ is a factor of $6x^{2} + 9x$ because $6x^{2} + 9x = 3x(2x + 3)$ .
Factorise
Examiner keyword
Rewrite an expression as a product of factors. Reverse of expanding.
Example
$x^{2} + 5x + 6 = (x + 2)(x + 3)$ .
Fully factorise
Examiner keyword
Continue factorising until no more factorising is possible. Apply all relevant techniques.
Example
$2x^{2} - 8 = 2(x^{2} - 4) = 2(x - 2)(x + 2)$ .

Common Mistakes and Misconceptions — Factorising

The traps other students keep falling into on factorising questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Stopping after one round of factorising
4MA1/1H — examiner reports 2022-2024
Why it happens
Student factors out a common factor and forgets to check for more.
How to avoid it
After EACH step, ask: 'can this factor further?' Difference of squares? Quadratic pattern? Common factor? Stop only when ALL pieces are irreducible.
✕Trying to factorise $x^{2} + 9$ as $(x + 3)(x + 3)$
Why it happens
Confusing sum of squares with difference.
How to avoid it
$a^{2} + b^{2}$ DOESN'T factor over the reals. Only $a^{2} - b^{2} = (a-b)(a+b)$ . Test: $(x+3)(x+3) = x^{2} + 6x + 9 \ne x^{2} + 9$ .
✕Pulling out only PART of the common factor
Why it happens
Easier to spot small factors.
How to avoid it
Find the LARGEST common factor: HCF of all coefficients × HCF of all variable parts. $6x^{2} + 9x \to$ HCF $= 3x$ , not $3$ or $x$ .
✕Wrong signs in quadratic factorising
Why it happens
Not testing the result.
How to avoid it
After factoring, MULTIPLY OUT to verify. $(x - 2)(x + 7)$ ? Expand: $x^{2} + 7x - 2x - 14 = x^{2} + 5x - 14$ . ✓

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

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Factorising — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Factorising

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Factorise out a common factor

2Difference of two squares

3Factorise a quadratic with leading coefficient 1

4Factorise a quadratic with leading coefficient > 1

5Factorise fully — common factor + difference of squares

Difference of two squares

Factor (algebra)

Factorise

Fully factorise

✕Stopping after one round of factorising

✕Trying to factorise x2+9x^{2} + 9x2+9 as (x+3)(x+3)(x + 3)(x + 3)(x+3)(x+3)

✕Pulling out only PART of the common factor

✕Wrong signs in quadratic factorising

Practice questions

Past paper style quiz

Assess your understanding

Factorising — frequently asked questions

✕Trying to factorise $x^{2} + 9$ as $(x + 3)(x + 3)$