Factorising · Study Notes · Edexcel IGCSE 4MA1 Higher

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

At a glance

Common factor first: pull out HCF.
Difference of squares: $a^{2} - b^{2} = (a-b)(a+b)$ .
Quadratic, $a = 1$ : find pair multiplying to $c$ , adding to $b$ .
Quadratic, $a > 1$ : AC method (split middle term).
'Fully factorise': apply ALL techniques in turn.
Always TEST by expanding back.

What you’ll learn

Mapped to the Pearson Edexcel 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

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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

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$a^{2} - b^{2} = (a-b)(a+b)$ . 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:

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

Cutting a small square from a large one leaves an L-shape that rearranges into a rectangle of sides (a − b) and (a + b).

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 $a = 1$

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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: $(x + 2)(x + 3)$ .

The right pair must multiply to c and add to b — here 2 and 3 give brackets (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

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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

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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.

Quick recap

Common factor first.
Difference of squares: $a^{2} - b^{2} = (a-b)(a+b)$ .
Quadratic $a=1$ : pair multiplies to $c$ , adds to $b$ .
Quadratic $a>1$ : AC method.
Fully factorise: apply ALL techniques.
Test by expanding back.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$a^{2} - b^{2} = (a-b)(a+b)$
Quadratic factoring: pair multiplies to product, adds to sum.
AC method: $p \times q = ac$ , $p + q = b$ .
Always check for common factor first.
Test by expanding to verify.

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.

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
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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
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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
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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
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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
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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.
6Recognising a perfect square [1 mark]
Getting started• Paper 1H/2H style — 1 mark• factorising, perfect square
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Question
Factorise $x^{2} - 18x + 81$ [1]
Step-by-step solution
1. Step 1
  Look for two numbers that multiply to $81$ and add to $-18$ . They are $-9$ and $-9$ , the same number twice, so the quadratic is a perfect square.
2. Step 2
  B1 Write it as a squared bracket.
  $x^{2} - 18x + 81 = (x - 9)(x - 9) = (x - 9)^{2}$
Answer
$(x - 9)^{2}$
Examiner tip
Examiner tip: A quadratic whose first and last terms are both squares, with a middle term equal to twice their square roots multiplied together ( $2 \times x \times 9 = 18x$ ), is a perfect square. Either $(x - 9)^{2}$ or $(x - 9)(x - 9)$ scores the mark.

Common mistake: Writing $(x - 9)(x + 9)$ . That is $x^{2} - 81$ , a difference of two squares, with no middle term at all.
7Factorising a formula fully [2 marks]
Getting started• Paper 1H/2H style — 2 marks• factorising, common factor, formulae
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Question
The total surface area of a cylinder is $\pi r^{2} + 2\pi rh + \pi r^{2}$ , which simplifies to $2\pi r^{2} + 2\pi rh$ .

Factorise fully $2\pi r^{2} + 2\pi rh$ [2]
Step-by-step solution
1. Step 1
  Treat $\pi$ like any other letter. Both terms contain $2$ , $\pi$ and $r$ , so the HCF is $2\pi r$ .
2. Step 2
  B2 Divide each term by $2\pi r$ .
  $2\pi r^{2} + 2\pi rh = 2\pi r(r + h)$
Answer
$2\pi r(r + h)$
Examiner tip
Examiner tip: $\pi$ is a number, but in algebra it is treated as a common factor just like a letter. Check by expanding: $2\pi r \times r = 2\pi r^{2}$ and $2\pi r \times h = 2\pi rh$ .

Common mistake: Taking out only part of the HCF, as in $2\pi(r^{2} + rh)$ or $2r(\pi r + \pi h)$ . These score 1 of the 2 marks because the bracket still has a common factor.
8Grouping a cubic, then difference of two squares [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• factorising, grouping, difference of two squares
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Question
Factorise fully $x^{3} + 2x^{2} - 9x - 18$ [3]
Step-by-step solution
1. Step 1
  M1 Group the terms in pairs. Take $x^{2}$ out of the first pair and $-9$ out of the second.
  $x^{3} + 2x^{2} - 9x - 18 = x^{2}(x + 2) - 9(x + 2)$
2. Step 2
  A1 Take out the common bracket $(x + 2)$ .
  $= (x + 2)(x^{2} - 9)$
3. Step 3
  A1 $x^{2} - 9$ is a difference of two squares, so split it.
  $= (x + 2)(x - 3)(x + 3)$
Answer
$(x + 2)(x - 3)(x + 3)$
Examiner tip
Examiner tip: After every factorisation, look at each bracket and ask whether it will factorise again. "Fully" is the clue that at least one more step is needed here.

Common mistake: Taking out $+9$ instead of $-9$ from the second pair, giving $x^{2}(x + 2) + 9(-x - 2)$ , which does not produce a matching bracket. Stopping at $(x + 2)(x^{2} - 9)$ scores 2 of the 3 marks.
9Non-monic quadratic, then a hence part with extra factors [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• factorising, non-monic quadratic, common factor
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Question
(a) Factorise $3x^{2} + 10x - 8$ [2]

(b) Hence factorise fully $3x^{3}y + 10x^{2}y - 8xy$ [2]
Step-by-step solution
1. Step 1
  M1 (a) $3 \times (-8) = -24$ . The numbers with product $-24$ and sum $10$ are $12$ and $-2$ . Split the middle term and group.
  $3x^{2} + 12x - 2x - 8 = 3x(x + 4) - 2(x + 4)$
2. Step 2
  A1 Take out the common bracket.
  $3x^{2} + 10x - 8 = (x + 4)(3x - 2)$
3. Step 3
  M1 (b) Every term contains $x$ and $y$ . Take out $xy$ and the bracket left is the quadratic from part (a).
  $3x^{3}y + 10x^{2}y - 8xy = xy(3x^{2} + 10x - 8)$
4. Step 4
  A1 Use part (a).
  $= xy(x + 4)(3x - 2)$
Answer
(a) $(x + 4)(3x - 2)$

(b) $xy(x + 4)(3x - 2)$
Examiner tip
Examiner tip: "Hence" means part (b) is built from part (a). Once $xy$ is taken out, the quadratic is identical, so no new factorising is needed.

Common mistake: In (a), writing $(3x + 4)(x - 2)$ , which has the right end terms but a middle term of $-2x$ . In (b), taking out only $x$ and leaving $y$ inside each term loses the final mark.
10Factorising by grouping, then evaluating [5 marks]
Stretch• Paper 1H/2H style — 5 marks• factorising, grouping, difference of two squares
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Question
(a) Factorise $x^{2} - y^{2} + 5x - 5y$ [3]

(b) Use your answer to part (a) to work out the value of $x^{2} - y^{2} + 5x - 5y$ when $x = 8.7$ and $y = 3.7$ . Show your working clearly. [2]
Step-by-step solution
1. Step 1
  M1 (a) Group the two squares together and the two linear terms together.
  $(x^{2} - y^{2}) + (5x - 5y)$
2. Step 2
  M1 Factorise each group. The first is a difference of two squares; the second has a common factor of $5$ .
  $(x - y)(x + y) + 5(x - y)$
3. Step 3
  A1 Take out the common bracket $(x - y)$ .
  $= (x - y)(x + y + 5)$
4. Step 4
  M1 (b) Substitute into the factorised form.
  $(8.7 - 3.7)(8.7 + 3.7 + 5) = 5 \times 17.4$
5. Step 5
  A1 Multiply.
  $5 \times 17.4 = 87$
Answer
(a) $(x - y)(x + y + 5)$

(b) $87$
Examiner tip
Examiner tip: Pairing the terms in the order printed ( $x^{2} - y^{2}$ and $5x - 5y$ ) works here, but it is the grouping by type that makes it work. If one grouping fails, try another.

Common mistake: Writing $(x - y)(x + y) + 5(x - y) = (x - y)(x + y)(5)$ . The common bracket is taken out once, and what is left, $x + y$ and $5$ , is added inside the second bracket. In (b), a bare $87$ from the calculator scores 0, because the question tells you to use part (a) and to show your working.

Question 1
Paper 1H/2H style — 1 mark1 mark
Factorise $3a(b + 5) - 2(b + 5)$ [1]
Model answer
The bracket $(b + 5)$ is common to both terms.

Answer: $(b + 5)(3a - 2)$
Why this scores
Mark scheme:
- B1 for $(b + 5)(3a - 2)$ oe
Why this scores full marks: The whole bracket $(b + 5)$ is taken out once, like a single letter. What is left from each term, $3a$ and $-2$ , goes into a new bracket with its sign. Answers such as $(b + 5)(-6a)$ or $(b + 5)^{2}(3a - 2)$ score 0.
Question 2
Paper 1H/2H style — 2 marks2 marks
Factorise $x^{2} - 15x + 56$ [2]
Model answer
Product $+56$ , sum $-15$ , so both numbers are negative: $-7 \times -8 = 56$ and $-7 + (-8) = -15$

Answer: $(x - 7)(x - 8)$
Why this scores
Mark scheme:
- M1 for $(x \pm 7)(x \pm 8)$
- A1 for $(x - 7)(x - 8)$ cao
Why this scores full marks: The script reasons about signs first: a positive product with a negative sum means both numbers are negative. That rules out half the pairs before any searching.
Question 3
Paper 1H/2H style — 2 marks2 marks
Use factorisation to work out $6.35^{2} - 3.65^{2}$

Show your working clearly. [2]
Model answer
$6.35^{2} - 3.65^{2} = (6.35 - 3.65)(6.35 + 3.65)$

$= 2.7 \times 10$

Answer: $27$
Why this scores
Mark scheme:
- M1 for $(6.35 - 3.65)(6.35 + 3.65)$ or $2.7 \times 10$
- A1 for $27$ cao
A correct answer with no working scores 0.

Why this scores full marks: The difference of two squares turns two awkward squares into a subtraction and an addition. The sum is exactly $10$ , so the final product is easy to do without a calculator.
Question 4
Paper 1H/2H style — 2 marks2 marks
$n$ is an integer.

By factorising, show that $n^{2} + 3n$ is always even. [2]
Model answer
$n^{2} + 3n = n(n + 3)$

If $n$ is even, the factor $n$ is even, so the product is even.

If $n$ is odd, then $n + 3$ is odd + odd, which is even, so the product is even.

So $n^{2} + 3n$ is always even.
Why this scores
Mark scheme:
- M1 for $n(n + 3)$
- A1 for a complete argument covering both $n$ even and $n$ odd
Why this scores full marks: Once the expression is a product, it is even as soon as one factor is even. The script checks both cases, so no integer is missed. Testing a few values, such as $n = 1, 2, 3$ , is not a proof.
Question 5
Paper 1H/2H style — 3 marks3 marks
Factorise fully $6(x + 2)^{2} - 9(x + 2)$ [3]
Model answer
The HCF of the two terms is $3(x + 2)$ .

$6(x + 2)^{2} - 9(x + 2) = 3(x + 2)\big[2(x + 2) - 3\big]$

$= 3(x + 2)(2x + 4 - 3)$

Answer: $3(x + 2)(2x + 1)$
Why this scores
Mark scheme:
- M1 for $(x + 2)\big[6(x + 2) - 9\big]$ or $3(x + 2)\big[2(x + 2) - 3\big]$ oe, or for expanding correctly to $6x^{2} + 15x + 6$
- M1 for simplifying the second bracket to $2x + 1$ or $6x + 3$
- A1 for $3(x + 2)(2x + 1)$ oe
Why this scores full marks: The script takes out the number $3$ and the bracket $(x + 2)$ together, then simplifies what is left. $(x + 2)(6x + 3)$ is correct but not fully factorised, because $6x + 3 = 3(2x + 1)$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
Factorise $6x^{2} + 5xy - 6y^{2}$ [3]
Model answer
$6 \times (-6) = -36$ ; $9 \times (-4) = -36$ and $9 + (-4) = 5$

$6x^{2} + 9xy - 4xy - 6y^{2}$

$= 3x(2x + 3y) - 2y(2x + 3y)$

Answer: $(2x + 3y)(3x - 2y)$
Why this scores
Mark scheme:
- M1 for splitting the middle term as $9xy - 4xy$ oe
- M1 for $3x(2x + 3y) - 2y(2x + 3y)$ oe, or $(2x \pm 3y)(3x \pm 2y)$
- A1 for $(2x + 3y)(3x - 2y)$ cao
Why this scores full marks: The script uses the $ac$ method with $-36$ , then groups in pairs so the common bracket appears. The $y$ s ride along with the numbers, since $3y \times (-2y) = -6y^{2}$ .
Question 7
Paper 1H/2H style — 3 marks3 marks
$a$ and $b$ are numbers such that

$a^{2} - b^{2} = 45$ and $a - b = 3$

Find the value of $a$ and the value of $b$ . [3]
Model answer
$a^{2} - b^{2} = (a - b)(a + b)$

$3(a + b) = 45$ , so $a + b = 15$

Adding $a + b = 15$ and $a - b = 3$ : $2a = 18$

Answer: $a = 9$ , $b = 6$
Why this scores
Mark scheme:
- M1 for $(a - b)(a + b) = 45$ oe
- M1 for $a + b = 15$
- A1 for $a = 9$ and $b = 6$
Why this scores full marks: Factorising the difference of two squares turns one hard equation into two easy linear ones. Check: $81 - 36 = 45$ and $9 - 6 = 3$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
Factorise fully $6x^{3}y + 3x^{2}y^{2} - 45xy^{3}$ [4]
Model answer
The HCF of the three terms is $3xy$ .

$6x^{3}y + 3x^{2}y^{2} - 45xy^{3} = 3xy(2x^{2} + xy - 15y^{2})$

$2 \times (-15) = -30$ ; $6 \times (-5) = -30$ and $6 + (-5) = 1$

$2x^{2} + 6xy - 5xy - 15y^{2} = 2x(x + 3y) - 5y(x + 3y) = (x + 3y)(2x - 5y)$

Answer: $3xy(x + 3y)(2x - 5y)$
Why this scores
Mark scheme:
- B1 for $3xy(2x^{2} + xy - 15y^{2})$
- M1 for $(x \pm 3y)(2x \pm 5y)$ or $2x(x + 3y) - 5y(x + 3y)$ oe
- A1 for $(x + 3y)(2x - 5y)$
- A1 for $3xy(x + 3y)(2x - 5y)$ cao
Why this scores full marks: The script takes out the whole HCF, $3xy$ , before anything else, which leaves a quadratic in $x$ and $y$ with small coefficients. It then splits the middle term using $ac = -30$ , and the $y$ s ride along with the numbers. Leaving $3xy$ off the final line, or stopping at $3xy(2x^{2} + xy - 15y^{2})$ , loses the last mark.
Question 9
Paper 1H/2H style — 4 marks4 marks
Factorise fully $(x^{2} + 4x)^{2} - 25$ [4]
Model answer
$(x^{2} + 4x)^{2} - 5^{2} = (x^{2} + 4x - 5)(x^{2} + 4x + 5)$

$x^{2} + 4x - 5 = (x + 5)(x - 1)$

$x^{2} + 4x + 5$ : no two integers multiply to $5$ and add to $4$ , so it does not factorise.

Answer: $(x + 5)(x - 1)(x^{2} + 4x + 5)$
Why this scores
Mark scheme:
- M1 for $(x^{2} + 4x - 5)(x^{2} + 4x + 5)$
- A1 for both brackets correct
- M1 for $(x + 5)(x - 1)$
- A1 for $(x + 5)(x - 1)(x^{2} + 4x + 5)$ cao
Why this scores full marks: The script keeps $x^{2} + 4x$ as one block for the difference of two squares, rather than expanding to a quartic. It then factorises the bracket that can be factorised and says why the other cannot.
Question 10
Paper 1H/2H style — 5 marks5 marks
(a) Factorise $2x^{2} - 7x - 15$ [2]

(b) Use your answer to part (a) to explain why $2n^{2} - 7n - 15$ is not a prime number for any integer $n > 6$ [3]
Model answer
(a) $2 \times (-15) = -30$ ; $-10 \times 3 = -30$ and $-10 + 3 = -7$

$2x^{2} - 10x + 3x - 15 = 2x(x - 5) + 3(x - 5)$

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

(b) $2n^{2} - 7n - 15 = (n - 5)(2n + 3)$

For an integer $n > 6$ , $n \geqslant 7$ , so $n - 5 \geqslant 2$ and $2n + 3 \geqslant 17$ .

So $2n^{2} - 7n - 15$ is the product of two integers that are both greater than $1$ .

A prime has only the factors $1$ and itself, so $2n^{2} - 7n - 15$ is not prime.
Why this scores
Mark scheme:
- M1 for $(x \pm 5)(2x \pm 3)$
- A1 for $(x - 5)(2x + 3)$ cao
- M1 for writing $2n^{2} - 7n - 15 = (n - 5)(2n + 3)$
- M1 for showing both factors are greater than $1$ when $n > 6$ , eg $n - 5 \geqslant 2$
- A1 for a complete explanation referring to the definition of a prime
Why this scores full marks: The factorised form shows the number as a product. The script then checks that neither factor can be $1$ , which is why $n > 6$ matters: at $n = 6$ the value is $1 \times 15$ , so the first factor is $1$ and this argument no longer applies.

4MA1 Factorising

At a glance

What you’ll learn

Common factor — always start here

Difference of two squares

Factorising quadratics with $a = 1$

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

Fully factorise — applying multiple techniques

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

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

6Recognising a perfect square [1 mark]

7Factorising a formula fully [2 marks]

8Grouping a cubic, then difference of two squares [3 marks]

9Non-monic quadratic, then a hence part with extra factors [4 marks]

10Factorising by grouping, then evaluating [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

4MA1 Factorising

At a glance

What you’ll learn

Common factor — always start here

Difference of two squares

Factorising quadratics with a=1a = 1a=1

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

Fully factorise — applying multiple techniques

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

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

6Recognising a perfect square [1 mark]

7Factorising a formula fully [2 marks]

8Grouping a cubic, then difference of two squares [3 marks]

9Non-monic quadratic, then a hence part with extra factors [4 marks]

10Factorising by grouping, then evaluating [5 marks]

Model AnswersFull mark-scheme-style answers

Factorising quadratics with $a = 1$

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