What is factorisation in algebra and why is it important in the Cambridge IGCSE Mathematics curriculum?

Factorisation in algebra involves breaking down an expression into a product of simpler expressions, or factors, that when multiplied together give the original expression. It is important in the Cambridge IGCSE Mathematics curriculum because it simplifies complex algebraic expressions, making it easier to solve equations and understand mathematical relationships.

How do you factorise a quadratic expression in the form ax^2 + bx + c?

To factorise a quadratic expression like ax^2 + bx + c, you need to find two numbers that multiply to ac (the product of the coefficient of x^2 and the constant term) and add up to b (the coefficient of x). Once these numbers are identified, you can rewrite the middle term and factor by grouping. This method is essential for solving quadratic equations in the Cambridge IGCSE Mathematics syllabus.

What are the common techniques used in factorisation of algebraic expressions?

Common techniques for factorisation include taking out the greatest common factor (GCF), using the difference of squares, factorising quadratics, and applying the sum or difference of cubes. Mastery of these techniques is crucial for students studying Cambridge IGCSE Mathematics as they form the foundation for more advanced algebraic manipulations.

Can you explain the difference between factorisation and simplification in algebra?

Factorisation involves rewriting an expression as a product of its factors, while simplification involves reducing an expression to its simplest form. Although both processes aim to make expressions easier to work with, factorisation specifically focuses on expressing the original equation as a multiplication of simpler terms. Understanding this distinction is important for Cambridge IGCSE Mathematics students to correctly approach algebraic problems.

How can factorisation help in solving algebraic equations?

Factorisation helps in solving algebraic equations by transforming complex expressions into simpler ones that can be easily solved. For example, once a quadratic equation is factorised, it can be solved by setting each factor equal to zero. This method is a key skill in the Cambridge IGCSE Mathematics curriculum, as it provides a systematic approach to finding solutions to equations.

Why is "Stopping before the factorisation is fully done" a common mistake in Factorisation?

Why it happens: Students take out a partial common factor and stop, e.g. 3(4x - 6y) instead of 6(2x - 3y). How to avoid it: After factoring, scan the bracket — if there's still a common factor, take it out. Source: 0580/42 — examiner reports each series

Why is "Trying to factorise a sum of two squares as if it were a difference" a common mistake in Factorisation?

Why it happens: a^2 + b^2 does NOT factor over the reals; only a^2 - b^2 does. How to avoid it: Difference of two squares only — never sum of two squares.

Why is "Wrong signs in the quadratic brackets" a common mistake in Factorisation?

Why it happens: Students find the right pair of numbers but assign the signs incorrectly. How to avoid it: Always re-expand to check: (x + p)(x + q) = x^2 + (p+q)x + pq. Match coefficients.

Why is "Trying to factor $ax^{2} + bx + c$ (with $a \neq 1$) by inspection" a common mistake in Factorisation?

Why it happens: Inspection works for a=1; for a ≠ 1 it usually doesn't. How to avoid it: Use the splitting method: find two numbers whose product is ac and sum is b. Source: 0580 Extended — recurring

Cambridge IGCSE Mathematics (0580)

Factorisation

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Short Notes - Factorisation

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Factorisation — Cambridge IGCSE 0580 Maths Extended (2026)

The reverse of expanding. Pull out common factors, recognise difference of two squares, factorise quadratics and group four-term expressions. The skill that unlocks solving quadratics, simplifying algebraic fractions and surds.

At a glance

Factorise = write as a PRODUCT of expressions.
Step 1 always: pull out the highest common factor (HCF).
Difference of two squares: $a^{2} - b^{2} = (a + b)(a - b)$ .
Quadratic $x^{2} + bx + c$ : find two numbers that ADD to $b$ and MULTIPLY to $c$ .
Quadratic $ax^{2} + bx + c$ : split-the-middle or AC method.
Four terms: factorise by grouping in pairs.
Always check by expanding back out.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.5 — Factorise where possible expressions of the form $ax + bx + kay + kby$ , $a^{2}x^{2} - b^{2}y^{2}$ , $a^{2} + 2ab + b^{2}$ , $ax^{2} + bx + c$ .

Step 1 — pull out the highest common factor

Always the first move. Find the HCF of every term and factor it out.

Before any other technique, look for what's common to every term: a number, a letter, or both.

Worked. Factorise $6x + 9$ .

HCF of $6$ and $9$ is $3$ .
$6x + 9 = 3(2x + 3)$ .

Worked. Factorise $4x^{2} - 12x$ .

HCF of $4$ and $12$ is $4$ . Both terms have at least one $x$ , so $x$ is also common.
$4x^{2} - 12x = 4x(x - 3)$ .

Both terms share 4 and one x, so 4x comes out; dividing each term by 4x gives what stays inside.

Worked. Factorise $15a^{3}b - 9a^{2}b^{2} + 6ab$ .

Numbers: HCF of $15, 9, 6$ is $3$ .
Letters: $a$ appears at least once everywhere; $b$ appears at least once everywhere.
HCF: $3ab$ .
$15a^{3}b - 9a^{2}b^{2} + 6ab = 3ab(5a^{2} - 3ab + 2)$ .

Always check by expanding the result — it should match the original.

Find the HCF of NUMBERS (use prime factorisation).
Find the HCF of LETTERS (lowest power present in every term).
Factor out the HCF.
Check by re-expanding.

Difference of two squares

If you see $a^{2} - b^{2}$ , factorise to $(a + b)(a - b)$ instantly.

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

Spot the signature:

TWO terms.
Subtraction.
Both terms are PERFECT SQUARES.

Worked. Factorise $x^{2} - 25$ .

Both are squares: $x^{2}$ and $5^{2}$ .
$= (x + 5)(x - 5)$ .

Worked. Factorise $9x^{2} - 16$ .

$9x^{2} = (3x)^{2}$ and $16 = 4^{2}$ .
$= (3x + 4)(3x - 4)$ .

The L-shape left after cutting a small square from a big one rearranges into a single rectangle.

With a common factor first. Always pull out the HCF before checking for DOTS.

Worked. Factorise $8x^{2} - 32$ .

HCF: $8$ . Pull out: $8(x^{2} - 4)$ .
$x^{2} - 4$ is DOTS: $(x + 2)(x - 2)$ .
Final: $8(x + 2)(x - 2)$ .

$a^{2} - b^{2} = (a + b)(a - b)$ .
TWO terms, subtraction, both squares.
Pull out the common factor FIRST, then check for DOTS.
Sum of squares ( $a^{2} + b^{2}$ ) doesn't factorise (over reals).

Quadratic $x^{2} + bx + c$

Find two numbers that ADD to $b$ and MULTIPLY to $c$ .

For a monic quadratic ( $a = 1$ ): $x^{2} + bx + c = (x + p)(x + q) \text{ where } p + q = b \text{ and } pq = c.$

Worked. Factorise $x^{2} + 7x + 12$ .

Two numbers that add to $7$ and multiply to $12$ : try pairs of $12$ . $1 \times 12$ , $2 \times 6$ , $3 \times 4$ .
$3 + 4 = 7$ ✓.
$= (x + 3)(x + 4)$ .

Top is the product, bottom is the sum; the side numbers 3 and 4 are the factors.

The four areas sum to x² + 7x + 12; the rectangle's sides are the factors (x+3) and (x+4).

Worked. Factorise $x^{2} - 8x + 15$ .

Add to $-8$ , multiply to $15$ . Both negative since product positive and sum negative.
$-3 + -5 = -8$ , $-3 \times -5 = 15$ ✓.
$= (x - 3)(x - 5)$ .

Worked. Factorise $x^{2} + 2x - 15$ .

Add to $2$ , multiply to $-15$ . One positive, one negative.
$5 + (-3) = 2$ , $5 \times -3 = -15$ ✓.
$= (x + 5)(x - 3)$ .

Sign rules.

$c > 0$ : both factors have the SAME sign as $b$ .
$c < 0$ : the factors have OPPOSITE signs; the larger magnitude one matches the sign of $b$ .

Two numbers that add to $b$ and multiply to $c$ .
$c > 0$ : same signs.
$c < 0$ : opposite signs.
Always expand back out to verify.

Quadratic $ax^{2} + bx + c$ with $a \ne 1$

Multiply $a \cdot c$ , find two numbers that add to $b$ and multiply to that, then split-the-middle.

AC method (split-the-middle).

Compute $a \cdot c$ .
Find two numbers that ADD to $b$ and MULTIPLY to $a \cdot c$ .
Replace $bx$ with those two terms.
Factorise by grouping (pair the four terms).

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

$a \cdot c = 2 \times 3 = 6$ .
Add to $7$ , multiply to $6$ . $1$ and $6$ : $1 + 6 = 7$ ✓.
Split: $2x^{2} + x + 6x + 3$ .
Group: $x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$ .

Split the middle term using a·c, then group in pairs — both pairs leave the same bracket.

Worked. Factorise $3x^{2} - 10x + 8$ .

$a \cdot c = 24$ .
Add to $-10$ , multiply to $24$ . Both negative: $-4$ and $-6$ ✓.
Split: $3x^{2} - 4x - 6x + 8$ .
Group: $x(3x - 4) - 2(3x - 4) = (3x - 4)(x - 2)$ .

Tip. When grouping doesn't work, you've split the middle wrong — re-check signs.

Compute $a \cdot c$ .
Find two numbers: ADD to $b$ , MULTIPLY to $ac$ .
Split the middle term, then group in pairs.
Each pair shares a common factor.

Four-term expressions: factorise by grouping

Pair them up. Pull out the common factor from each pair. The remaining bracket should match.

Worked. Factorise $ax + bx + ay + by$ .

Group: $(ax + bx) + (ay + by) = x(a + b) + y(a + b)$ .
Common bracket: $(a + b)$ .
$= (a + b)(x + y)$ .

Worked. Factorise $6xy - 9y + 4x - 6$ .

Group: $(6xy - 9y) + (4x - 6) = 3y(2x - 3) + 2(2x - 3)$ .
Common bracket: $(2x - 3)$ .
$= (2x - 3)(3y + 2)$ .

If the bracket from each pair doesn't match, try regrouping. Sometimes you need to swap two adjacent terms.

Pair the four terms.
Pull out the common factor from each pair.
The two brackets must match — that's your common factor.
If they don't match, regroup.

Quick recap

Step 1 ALWAYS: pull out the HCF.
Difference of two squares: $a^{2} - b^{2} = (a+b)(a-b)$ .
Monic quadratic $x^{2} + bx + c$ : two numbers that add to $b$ and multiply to $c$ .
Non-monic $ax^{2} + bx + c$ : AC method (split the middle).
Four terms: factorise by grouping.
Always check by expanding back out.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Factorise — write an expression as a product of factors.
Highest common factor (HCF) — the largest factor shared by all terms.
Difference of two squares — $a^{2} - b^{2} = (a + b)(a - b)$ .
Quadratic expression — a polynomial of degree 2: $ax^{2} + bx + c$ .

How it’s examined

Factorisation appears on every paper. Paper 2 has 1-2 mark single-skill questions ('factorise $x^{2} - 49$ ', 'factorise $x^{2} + 7x + 10$ '). Paper 4 uses factorisation as a step inside solving quadratics or simplifying algebraic fractions (3-5 marks total). Examiner reports flag forgetting the HCF first and miscounting the sign on the middle term as recurring slips.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.5); 0580/22 May/Jun 2024 — Q12 (factorise quadratic); 0580/42 Oct/Nov 2024 — Q5 (factorise by grouping); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Factorisation

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

Question type:

Question patterns to master — Factorisation

Almost every factorisation exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — factorise or factorise fully — on one expression, with no equation to solve.

How to approach it

Always take out the highest common factor first, then identify the structure: difference of two squares ( $a^2-b^2$ ), a quadratic trinomial (find two numbers), grouping for four terms, or a perfect square. Re-expand to check.

Common trap

Stopping before the expression is fully factorised — $3(4x-6y)$ instead of $6(2x-3y)$ . Examiner reports note 'factorise fully' is code for: scan the bracket for any remaining common factor.

Multi-step problem

Recognise it by

Two linked parts — factorise … then hence solve … — where the second part must reuse the first.

How to approach it

Factorise fully in part (a), then in part (b) set each factor equal to zero (a constant factor never gives a root) and read off the solutions.

Common trap

Ignoring the word hence and solving the equation by a fresh method. The mark scheme credits only the factorised-form route once hence is stated.

1Take out the highest common factor
CoreDirect calculation• common factor
Question
Factorise fully $12x^{2} y - 18 x y^{2}$ .
Step-by-step solution
1. Step 1
  Identify the largest factor common to both terms: HCF of $12$ and $18$ is $6$ , then take $xy$ .
  $\text{Common factor} = 6xy$
2. Step 2
  Divide each term by the common factor.
  $\dfrac{12x^{2}y}{6xy} = 2x,\quad \dfrac{18xy^{2}}{6xy} = 3y$
3. Step 3
  Write as common factor times bracket.
  $12x^{2}y - 18xy^{2} = 6xy(2x - 3y)$
Answer
$6xy(2x - 3y)$
Examiner tip
"Factorise fully" is examiner code for: don't stop until no further common factor exists. $3xy(4x - 6y)$ scores 1/2 because $4x - 6y$ has another factor of $2$ .
2Factorise a difference of two squares
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2023 Q14• difference of two squares
Question
Factorise $25x^{2} - 49$ .
Step-by-step solution
1. Step 1
  Recognise both terms as squares.
  $25x^{2} = (5x)^{2},\quad 49 = 7^{2}$
2. Step 2
  Apply $a^{2} - b^{2} = (a + b)(a - b)$ .
  $(5x + 7)(5x - 7)$
Answer
$(5x + 7)(5x - 7)$
3Factorise a quadratic trinomial
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q5• quadratic
Question
Factorise $x^{2} - 7x + 12$ .
Step-by-step solution
1. Step 1
  Find two numbers whose product is $12$ and sum is $-7$ : $-3$ and $-4$ .
2. Step 2
  Write as two brackets.
  $(x - 3)(x - 4)$
Answer
$(x - 3)(x - 4)$
Examiner tip
Always check by re-expanding: $(x-3)(x-4) = x^{2} - 4x - 3x + 12 = x^{2} - 7x + 12$ . ✓
4Factorise by grouping
ExtendedDirect calculation• grouping
Question
Factorise $ax + ay + bx + by$ .
Step-by-step solution
1. Step 1
  Group the first two and last two terms.
  $(ax + ay) + (bx + by)$
2. Step 2
  Factor each group.
  $= a(x + y) + b(x + y)$
3. Step 3
  Factor the common bracket.
  $= (x + y)(a + b)$
Answer
$(x + y)(a + b)$
5Factorise $2x^{2} + 7x + 3$
ExtendedDirect calculation• quadratic, coefficient
Question
Factorise $2x^{2} + 7x + 3$ .
Step-by-step solution
1. Step 1
  Multiply $a \cdot c = 2 \cdot 3 = 6$ . Find two numbers that multiply to $6$ and add to $7$ : $1$ and $6$ .
2. Step 2
  Split the middle term.
  $2x^{2} + 1x + 6x + 3$
3. Step 3
  Group and factor.
  $x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$
Answer
$(2x + 1)(x + 3)$
6Take out a numeric common factor
CoreDirect calculation• common factor
Question
Factorise $15x + 25$ .
Step-by-step solution
1. Step 1
  The HCF of $15$ and $25$ is $5$ .
2. Step 2
  Divide each term by $5$ and write as a product.
  $15x + 25 = 5(3x + 5)$
Answer
$5(3x + 5)$
Examiner tip
Always check the bracket has no further common factor. $5(3x+5)$ is fully factorised because $\gcd(3, 5) = 1$ .
7Difference of two squares with a leading coefficient
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2024 Q12• difference of two squares, common factor
Question
Factorise fully $18x^{2} - 50$ .
Step-by-step solution
1. Step 1
  Take out the common factor $2$ first.
  $18x^{2} - 50 = 2(9x^{2} - 25)$
2. Step 2
  Recognise $9x^{2} - 25$ as a difference of two squares: $(3x)^{2} - 5^{2}$ .
3. Step 3
  Apply $a^{2} - b^{2} = (a + b)(a - b)$ .
  $= 2(3x + 5)(3x - 5)$
Answer
$2(3x + 5)(3x - 5)$
Examiner tip
The 2024 mark scheme awards method marks for taking the common factor out first. Jumping straight to $(\sqrt{18}\,x)^{2} - (\sqrt{50})^{2}$ leaves surds and loses the accuracy mark.
8Factorise a quadratic with a negative constant
ExtendedDirect calculation• Adapted from 0580/22 Feb/Mar 2024 Q10• quadratic, signs
Question
Factorise $x^{2} + 2x - 15$ .
Step-by-step solution
1. Step 1
  Find two numbers whose product is $-15$ and sum is $+2$ . Try $5$ and $-3$ : $5 \times (-3) = -15$ , $5 + (-3) = 2$ . ✓
2. Step 2
  Write as two brackets.
  $(x + 5)(x - 3)$
Answer
$(x + 5)(x - 3)$
Examiner tip
The examiner report flags candidates often swap the signs and write $(x - 5)(x + 3)$ , which expands to $x^{2} - 2x - 15$ . Re-expand to verify the middle term matches.
9Four-term grouping with a sign change
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q13• grouping, signs
Question
Factorise $2xy - 6x - 5y + 15$ .
Step-by-step solution
1. Step 1
  Group the first two and last two terms.
  $(2xy - 6x) + (-5y + 15)$
2. Step 2
  Factor each group. Take out $-5$ from the second group so the bracket matches.
  $= 2x(y - 3) - 5(y - 3)$
3. Step 3
  Factor the common bracket.
  $= (y - 3)(2x - 5)$
Answer
$(y - 3)(2x - 5)$
Examiner tip
The examiner report flags candidates often take out $+5$ from $-5y + 15$ and get $(y + 3)$ — wrong sign on the inner bracket. Taking out $-5$ matches the bracket from the first group.
10Factorise a perfect-square trinomial
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q9• perfect square, trinomial
Question
Factorise $9x^{2} - 24x + 16$ .
Step-by-step solution
1. Step 1
  Check whether the first and last terms are perfect squares: $9x^{2} = (3x)^{2}$ and $16 = 4^{2}$ .
2. Step 2
  Check the middle term: $2 \times 3x \times 4 = 24x$ . The sign is $-$ , so the bracket is $(3x - 4)$ .
3. Step 3
  Write as a square.
  $9x^{2} - 24x + 16 = (3x - 4)^{2}$
Answer
$(3x - 4)^{2}$
Examiner tip
Examiners reward candidates who recognise the perfect-square pattern and avoid the splitting method. Always check the middle term equals $2ab$ before writing the squared bracket.
11Factorise using a surd factor
ChallengeDirect calculation• Adapted from 0580/42 May/Jun 2024 Q19• surds, factorising
Question
Factorise $x^{2} - 7$ over the real numbers.
Step-by-step solution
1. Step 1
  Although $7$ is not a perfect square, $x^{2} - 7$ is still a difference of two squares: $x^{2} - (\sqrt{7})^{2}$ .
2. Step 2
  Apply $a^{2} - b^{2} = (a + b)(a - b)$ with $b = \sqrt{7}$ .
  $= (x + \sqrt{7})(x - \sqrt{7})$
Answer
$(x + \sqrt{7})(x - \sqrt{7})$
Examiner tip
The mark scheme accepts surd factors when the question asks to factorise over the reals. Leave the surds exact — converting $\sqrt{7}$ to a decimal loses the accuracy mark.
12Factorise using a common bracket
ChallengeDirect calculation• Adapted from 0580/42 Feb/Mar 2024 Q19• common factor, brackets
Question
Factorise fully $5(x + 2)^{2} - 3(x + 2)$ .
Step-by-step solution
1. Step 1
  Treat $(x + 2)$ as the common factor. Both terms contain it.
2. Step 2
  Take out the lowest power: $(x + 2)^{1}$ . The first term leaves $5(x + 2)$ ; the second leaves $-3$ .
  $5(x + 2)^{2} - 3(x + 2) = (x + 2)\,[5(x + 2) - 3]$
3. Step 3
  Simplify the inner bracket.
  $= (x + 2)(5x + 10 - 3) = (x + 2)(5x + 7)$
Answer
$(x + 2)(5x + 7)$
Examiner tip
A* candidates secure full marks by recognising the entire bracket $(x + 2)$ as a common factor. Expanding everything first and re-factorising the resulting trinomial is valid but wastes time and risks sign errors.
13Factorise an expression and use it to solve an equation
ChallengeMulti-step problem• Adapted from 0580/42 Oct/Nov 2024 Q18• factorising, solving, multi-step
Question
(a) Factorise $3x^{2} - 12$ . (b) Hence solve $3x^{2} - 12 = 0$ .
Step-by-step solution
1. Step 1
  (a) Take out the common factor $3$ first.
  $3x^{2} - 12 = 3(x^{2} - 4)$
2. Step 2
  Apply the difference of two squares to the bracket.
  $= 3(x + 2)(x - 2)$
3. Step 3
  (b) Set each factor equal to zero. The factor $3$ never gives zero, so use $(x + 2) = 0$ or $(x - 2) = 0$ .
  $x = -2 \ \text{or}\ x = 2$
Answer
(a) $3(x + 2)(x - 2)$ (b) $x = \pm 2$
Examiner tip
The word "hence" signals you must use part (a). The mark scheme awards no credit for solving (b) by a separate method (e.g. quadratic formula) — you must use the factorised form.

Key Formulae — Factorisation

The formulae you need to memorise for factorisation on the Cambridge IGCSE 0580 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 algebraic expressions
When to use
Whenever you spot a subtraction of two terms that are both perfect squares.
Quadratic factorisation (monic)
$x^{2} + (p+q)x + pq = (x + p)(x + q)$
$p, q$
two numbers whose sum is the middle coefficient and whose product is the constant term
When to use
Use whenever the leading coefficient is $1$ — find the two numbers and write the brackets.
Perfect-square trinomial
$a^{2} \pm 2ab + b^{2} = (a \pm b)^{2}$
When to use
Recognise this when the first and last terms are perfect squares and the middle term is twice the product of their roots.

Key Definitions and Keywords — Factorisation

Definitions to memorise and the exact keywords mark schemes credit for factorisation answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Factorise
Examiner keyword
Rewrite an algebraic expression as a product of two or more factors.
Factorise fully
Examiner keyword
Factorise until no further common factor or factorisable structure remains.
Highest common factor (HCF) of algebraic terms
Examiner keyword
The largest expression that divides each given term exactly — found by taking the lower power of each shared variable and the HCF of the coefficients.
Example
HCF of $12x^{2}y$ and $18xy^{2}$ is $6xy$ .
Quadratic trinomial
An expression of the form $ax^{2} + bx + c$ with $a \neq 0$ .
Difference of two squares
Examiner keyword
An expression of the form $a^{2} - b^{2}$ , which always factors as $(a+b)(a-b)$ .

Common Mistakes and Misconceptions — Factorisation

The traps other students keep falling into on factorisation questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Stopping before the factorisation is fully done
0580/42 — examiner reports each series
Why it happens
Students take out a partial common factor and stop, e.g. $3(4x - 6y)$ instead of $6(2x - 3y)$ .
How to avoid it
After factoring, scan the bracket — if there's still a common factor, take it out.
✕Trying to factorise a sum of two squares as if it were a difference
Why it happens
$a^{2} + b^{2}$ does NOT factor over the reals; only $a^{2} - b^{2}$ does.
How to avoid it
Difference of two squares only — never sum of two squares.
✕Wrong signs in the quadratic brackets
Why it happens
Students find the right pair of numbers but assign the signs incorrectly.
How to avoid it
Always re-expand to check: $(x + p)(x + q) = x^{2} + (p+q)x + pq$ . Match coefficients.
✕Trying to factor $ax^{2} + bx + c$ (with $a \neq 1$ ) by inspection
0580 Extended — recurring
Why it happens
Inspection works for $a=1$ ; for $a \neq 1$ it usually doesn't.
How to avoid it
Use the splitting method: find two numbers whose product is $ac$ and sum is $b$ .

Factorisation — frequently asked questions

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

Cambridge IGCSE Mathematics (0580)

Factorisation

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Short Notes - Factorisation

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Factorisation — Cambridge IGCSE 0580 Maths Extended (2026)

At a glance

Factorise = write as a PRODUCT of expressions.
Step 1 always: pull out the highest common factor (HCF).
Difference of two squares: $a^{2} - b^{2} = (a + b)(a - b)$ .
Quadratic $x^{2} + bx + c$ : find two numbers that ADD to $b$ and MULTIPLY to $c$ .
Quadratic $ax^{2} + bx + c$ : split-the-middle or AC method.
Four terms: factorise by grouping in pairs.
Always check by expanding back out.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.5 — Factorise where possible expressions of the form $ax + bx + kay + kby$ , $a^{2}x^{2} - b^{2}y^{2}$ , $a^{2} + 2ab + b^{2}$ , $ax^{2} + bx + c$ .

Step 1 — pull out the highest common factor

Always the first move. Find the HCF of every term and factor it out.

Before any other technique, look for what's common to every term: a number, a letter, or both.

Worked. Factorise $6x + 9$ .

HCF of $6$ and $9$ is $3$ .
$6x + 9 = 3(2x + 3)$ .

Worked. Factorise $4x^{2} - 12x$ .

HCF of $4$ and $12$ is $4$ . Both terms have at least one $x$ , so $x$ is also common.
$4x^{2} - 12x = 4x(x - 3)$ .

Both terms share 4 and one x, so 4x comes out; dividing each term by 4x gives what stays inside.

Worked. Factorise $15a^{3}b - 9a^{2}b^{2} + 6ab$ .

Numbers: HCF of $15, 9, 6$ is $3$ .
Letters: $a$ appears at least once everywhere; $b$ appears at least once everywhere.
HCF: $3ab$ .
$15a^{3}b - 9a^{2}b^{2} + 6ab = 3ab(5a^{2} - 3ab + 2)$ .

Always check by expanding the result — it should match the original.

Find the HCF of NUMBERS (use prime factorisation).
Find the HCF of LETTERS (lowest power present in every term).
Factor out the HCF.
Check by re-expanding.

Difference of two squares

If you see $a^{2} - b^{2}$ , factorise to $(a + b)(a - b)$ instantly.

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

Spot the signature:

TWO terms.
Subtraction.
Both terms are PERFECT SQUARES.

Worked. Factorise $x^{2} - 25$ .

Both are squares: $x^{2}$ and $5^{2}$ .
$= (x + 5)(x - 5)$ .

Worked. Factorise $9x^{2} - 16$ .

$9x^{2} = (3x)^{2}$ and $16 = 4^{2}$ .
$= (3x + 4)(3x - 4)$ .

The L-shape left after cutting a small square from a big one rearranges into a single rectangle.

With a common factor first. Always pull out the HCF before checking for DOTS.

Worked. Factorise $8x^{2} - 32$ .

HCF: $8$ . Pull out: $8(x^{2} - 4)$ .
$x^{2} - 4$ is DOTS: $(x + 2)(x - 2)$ .
Final: $8(x + 2)(x - 2)$ .

$a^{2} - b^{2} = (a + b)(a - b)$ .
TWO terms, subtraction, both squares.
Pull out the common factor FIRST, then check for DOTS.
Sum of squares ( $a^{2} + b^{2}$ ) doesn't factorise (over reals).

Quadratic $x^{2} + bx + c$

Find two numbers that ADD to $b$ and MULTIPLY to $c$ .

For a monic quadratic ( $a = 1$ ): $x^{2} + bx + c = (x + p)(x + q) \text{ where } p + q = b \text{ and } pq = c.$

Worked. Factorise $x^{2} + 7x + 12$ .

Two numbers that add to $7$ and multiply to $12$ : try pairs of $12$ . $1 \times 12$ , $2 \times 6$ , $3 \times 4$ .
$3 + 4 = 7$ ✓.
$= (x + 3)(x + 4)$ .

Top is the product, bottom is the sum; the side numbers 3 and 4 are the factors.

The four areas sum to x² + 7x + 12; the rectangle's sides are the factors (x+3) and (x+4).

Worked. Factorise $x^{2} - 8x + 15$ .

Add to $-8$ , multiply to $15$ . Both negative since product positive and sum negative.
$-3 + -5 = -8$ , $-3 \times -5 = 15$ ✓.
$= (x - 3)(x - 5)$ .

Worked. Factorise $x^{2} + 2x - 15$ .

Add to $2$ , multiply to $-15$ . One positive, one negative.
$5 + (-3) = 2$ , $5 \times -3 = -15$ ✓.
$= (x + 5)(x - 3)$ .

Sign rules.

$c > 0$ : both factors have the SAME sign as $b$ .
$c < 0$ : the factors have OPPOSITE signs; the larger magnitude one matches the sign of $b$ .

Two numbers that add to $b$ and multiply to $c$ .
$c > 0$ : same signs.
$c < 0$ : opposite signs.
Always expand back out to verify.

Quadratic $ax^{2} + bx + c$ with $a \ne 1$

Multiply $a \cdot c$ , find two numbers that add to $b$ and multiply to that, then split-the-middle.

AC method (split-the-middle).

Compute $a \cdot c$ .
Find two numbers that ADD to $b$ and MULTIPLY to $a \cdot c$ .
Replace $bx$ with those two terms.
Factorise by grouping (pair the four terms).

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

$a \cdot c = 2 \times 3 = 6$ .
Add to $7$ , multiply to $6$ . $1$ and $6$ : $1 + 6 = 7$ ✓.
Split: $2x^{2} + x + 6x + 3$ .
Group: $x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$ .

Split the middle term using a·c, then group in pairs — both pairs leave the same bracket.

Worked. Factorise $3x^{2} - 10x + 8$ .

$a \cdot c = 24$ .
Add to $-10$ , multiply to $24$ . Both negative: $-4$ and $-6$ ✓.
Split: $3x^{2} - 4x - 6x + 8$ .
Group: $x(3x - 4) - 2(3x - 4) = (3x - 4)(x - 2)$ .

Tip. When grouping doesn't work, you've split the middle wrong — re-check signs.

Compute $a \cdot c$ .
Find two numbers: ADD to $b$ , MULTIPLY to $ac$ .
Split the middle term, then group in pairs.
Each pair shares a common factor.

Four-term expressions: factorise by grouping

Pair them up. Pull out the common factor from each pair. The remaining bracket should match.

Worked. Factorise $ax + bx + ay + by$ .

Group: $(ax + bx) + (ay + by) = x(a + b) + y(a + b)$ .
Common bracket: $(a + b)$ .
$= (a + b)(x + y)$ .

Worked. Factorise $6xy - 9y + 4x - 6$ .

Group: $(6xy - 9y) + (4x - 6) = 3y(2x - 3) + 2(2x - 3)$ .
Common bracket: $(2x - 3)$ .
$= (2x - 3)(3y + 2)$ .

If the bracket from each pair doesn't match, try regrouping. Sometimes you need to swap two adjacent terms.

Pair the four terms.
Pull out the common factor from each pair.
The two brackets must match — that's your common factor.
If they don't match, regroup.

Quick recap

Step 1 ALWAYS: pull out the HCF.
Difference of two squares: $a^{2} - b^{2} = (a+b)(a-b)$ .
Monic quadratic $x^{2} + bx + c$ : two numbers that add to $b$ and multiply to $c$ .
Non-monic $ax^{2} + bx + c$ : AC method (split the middle).
Four terms: factorise by grouping.
Always check by expanding back out.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Factorise — write an expression as a product of factors.
Highest common factor (HCF) — the largest factor shared by all terms.
Difference of two squares — $a^{2} - b^{2} = (a + b)(a - b)$ .
Quadratic expression — a polynomial of degree 2: $ax^{2} + bx + c$ .

How it’s examined

Take this whole topic with you

Step-by-step worked examples — Factorisation

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

Question type:

Question patterns to master — Factorisation

Almost every factorisation exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — factorise or factorise fully — on one expression, with no equation to solve.

How to approach it

Common trap

Stopping before the expression is fully factorised — $3(4x-6y)$ instead of $6(2x-3y)$ . Examiner reports note 'factorise fully' is code for: scan the bracket for any remaining common factor.

Multi-step problem

Recognise it by

Two linked parts — factorise … then hence solve … — where the second part must reuse the first.

How to approach it

Factorise fully in part (a), then in part (b) set each factor equal to zero (a constant factor never gives a root) and read off the solutions.

Common trap

Ignoring the word hence and solving the equation by a fresh method. The mark scheme credits only the factorised-form route once hence is stated.

1Take out the highest common factor
CoreDirect calculation• common factor
Question
Factorise fully $12x^{2} y - 18 x y^{2}$ .
Step-by-step solution
1. Step 1
  Identify the largest factor common to both terms: HCF of $12$ and $18$ is $6$ , then take $xy$ .
  $\text{Common factor} = 6xy$
2. Step 2
  Divide each term by the common factor.
  $\dfrac{12x^{2}y}{6xy} = 2x,\quad \dfrac{18xy^{2}}{6xy} = 3y$
3. Step 3
  Write as common factor times bracket.
  $12x^{2}y - 18xy^{2} = 6xy(2x - 3y)$
Answer
$6xy(2x - 3y)$
Examiner tip
"Factorise fully" is examiner code for: don't stop until no further common factor exists. $3xy(4x - 6y)$ scores 1/2 because $4x - 6y$ has another factor of $2$ .
2Factorise a difference of two squares
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2023 Q14• difference of two squares
Question
Factorise $25x^{2} - 49$ .
Step-by-step solution
1. Step 1
  Recognise both terms as squares.
  $25x^{2} = (5x)^{2},\quad 49 = 7^{2}$
2. Step 2
  Apply $a^{2} - b^{2} = (a + b)(a - b)$ .
  $(5x + 7)(5x - 7)$
Answer
$(5x + 7)(5x - 7)$
3Factorise a quadratic trinomial
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q5• quadratic
Question
Factorise $x^{2} - 7x + 12$ .
Step-by-step solution
1. Step 1
  Find two numbers whose product is $12$ and sum is $-7$ : $-3$ and $-4$ .
2. Step 2
  Write as two brackets.
  $(x - 3)(x - 4)$
Answer
$(x - 3)(x - 4)$
Examiner tip
Always check by re-expanding: $(x-3)(x-4) = x^{2} - 4x - 3x + 12 = x^{2} - 7x + 12$ . ✓
4Factorise by grouping
ExtendedDirect calculation• grouping
Question
Factorise $ax + ay + bx + by$ .
Step-by-step solution
1. Step 1
  Group the first two and last two terms.
  $(ax + ay) + (bx + by)$
2. Step 2
  Factor each group.
  $= a(x + y) + b(x + y)$
3. Step 3
  Factor the common bracket.
  $= (x + y)(a + b)$
Answer
$(x + y)(a + b)$
5Factorise $2x^{2} + 7x + 3$
ExtendedDirect calculation• quadratic, coefficient
Question
Factorise $2x^{2} + 7x + 3$ .
Step-by-step solution
1. Step 1
  Multiply $a \cdot c = 2 \cdot 3 = 6$ . Find two numbers that multiply to $6$ and add to $7$ : $1$ and $6$ .
2. Step 2
  Split the middle term.
  $2x^{2} + 1x + 6x + 3$
3. Step 3
  Group and factor.
  $x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$
Answer
$(2x + 1)(x + 3)$
6Take out a numeric common factor
CoreDirect calculation• common factor
Question
Factorise $15x + 25$ .
Step-by-step solution
1. Step 1
  The HCF of $15$ and $25$ is $5$ .
2. Step 2
  Divide each term by $5$ and write as a product.
  $15x + 25 = 5(3x + 5)$
Answer
$5(3x + 5)$
Examiner tip
Always check the bracket has no further common factor. $5(3x+5)$ is fully factorised because $\gcd(3, 5) = 1$ .
7Difference of two squares with a leading coefficient
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2024 Q12• difference of two squares, common factor
Question
Factorise fully $18x^{2} - 50$ .
Step-by-step solution
1. Step 1
  Take out the common factor $2$ first.
  $18x^{2} - 50 = 2(9x^{2} - 25)$
2. Step 2
  Recognise $9x^{2} - 25$ as a difference of two squares: $(3x)^{2} - 5^{2}$ .
3. Step 3
  Apply $a^{2} - b^{2} = (a + b)(a - b)$ .
  $= 2(3x + 5)(3x - 5)$
Answer
$2(3x + 5)(3x - 5)$
Examiner tip
The 2024 mark scheme awards method marks for taking the common factor out first. Jumping straight to $(\sqrt{18}\,x)^{2} - (\sqrt{50})^{2}$ leaves surds and loses the accuracy mark.
8Factorise a quadratic with a negative constant
ExtendedDirect calculation• Adapted from 0580/22 Feb/Mar 2024 Q10• quadratic, signs
Question
Factorise $x^{2} + 2x - 15$ .
Step-by-step solution
1. Step 1
  Find two numbers whose product is $-15$ and sum is $+2$ . Try $5$ and $-3$ : $5 \times (-3) = -15$ , $5 + (-3) = 2$ . ✓
2. Step 2
  Write as two brackets.
  $(x + 5)(x - 3)$
Answer
$(x + 5)(x - 3)$
Examiner tip
The examiner report flags candidates often swap the signs and write $(x - 5)(x + 3)$ , which expands to $x^{2} - 2x - 15$ . Re-expand to verify the middle term matches.
9Four-term grouping with a sign change
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q13• grouping, signs
Question
Factorise $2xy - 6x - 5y + 15$ .
Step-by-step solution
1. Step 1
  Group the first two and last two terms.
  $(2xy - 6x) + (-5y + 15)$
2. Step 2
  Factor each group. Take out $-5$ from the second group so the bracket matches.
  $= 2x(y - 3) - 5(y - 3)$
3. Step 3
  Factor the common bracket.
  $= (y - 3)(2x - 5)$
Answer
$(y - 3)(2x - 5)$
Examiner tip
The examiner report flags candidates often take out $+5$ from $-5y + 15$ and get $(y + 3)$ — wrong sign on the inner bracket. Taking out $-5$ matches the bracket from the first group.
10Factorise a perfect-square trinomial
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q9• perfect square, trinomial
Question
Factorise $9x^{2} - 24x + 16$ .
Step-by-step solution
1. Step 1
  Check whether the first and last terms are perfect squares: $9x^{2} = (3x)^{2}$ and $16 = 4^{2}$ .
2. Step 2
  Check the middle term: $2 \times 3x \times 4 = 24x$ . The sign is $-$ , so the bracket is $(3x - 4)$ .
3. Step 3
  Write as a square.
  $9x^{2} - 24x + 16 = (3x - 4)^{2}$
Answer
$(3x - 4)^{2}$
Examiner tip
Examiners reward candidates who recognise the perfect-square pattern and avoid the splitting method. Always check the middle term equals $2ab$ before writing the squared bracket.
11Factorise using a surd factor
ChallengeDirect calculation• Adapted from 0580/42 May/Jun 2024 Q19• surds, factorising
Question
Factorise $x^{2} - 7$ over the real numbers.
Step-by-step solution
1. Step 1
  Although $7$ is not a perfect square, $x^{2} - 7$ is still a difference of two squares: $x^{2} - (\sqrt{7})^{2}$ .
2. Step 2
  Apply $a^{2} - b^{2} = (a + b)(a - b)$ with $b = \sqrt{7}$ .
  $= (x + \sqrt{7})(x - \sqrt{7})$
Answer
$(x + \sqrt{7})(x - \sqrt{7})$
Examiner tip
The mark scheme accepts surd factors when the question asks to factorise over the reals. Leave the surds exact — converting $\sqrt{7}$ to a decimal loses the accuracy mark.
12Factorise using a common bracket
ChallengeDirect calculation• Adapted from 0580/42 Feb/Mar 2024 Q19• common factor, brackets
Question
Factorise fully $5(x + 2)^{2} - 3(x + 2)$ .
Step-by-step solution
1. Step 1
  Treat $(x + 2)$ as the common factor. Both terms contain it.
2. Step 2
  Take out the lowest power: $(x + 2)^{1}$ . The first term leaves $5(x + 2)$ ; the second leaves $-3$ .
  $5(x + 2)^{2} - 3(x + 2) = (x + 2)\,[5(x + 2) - 3]$
3. Step 3
  Simplify the inner bracket.
  $= (x + 2)(5x + 10 - 3) = (x + 2)(5x + 7)$
Answer
$(x + 2)(5x + 7)$
Examiner tip
A* candidates secure full marks by recognising the entire bracket $(x + 2)$ as a common factor. Expanding everything first and re-factorising the resulting trinomial is valid but wastes time and risks sign errors.
13Factorise an expression and use it to solve an equation
ChallengeMulti-step problem• Adapted from 0580/42 Oct/Nov 2024 Q18• factorising, solving, multi-step
Question
(a) Factorise $3x^{2} - 12$ . (b) Hence solve $3x^{2} - 12 = 0$ .
Step-by-step solution
1. Step 1
  (a) Take out the common factor $3$ first.
  $3x^{2} - 12 = 3(x^{2} - 4)$
2. Step 2
  Apply the difference of two squares to the bracket.
  $= 3(x + 2)(x - 2)$
3. Step 3
  (b) Set each factor equal to zero. The factor $3$ never gives zero, so use $(x + 2) = 0$ or $(x - 2) = 0$ .
  $x = -2 \ \text{or}\ x = 2$
Answer
(a) $3(x + 2)(x - 2)$ (b) $x = \pm 2$
Examiner tip
The word "hence" signals you must use part (a). The mark scheme awards no credit for solving (b) by a separate method (e.g. quadratic formula) — you must use the factorised form.

Key Formulae — Factorisation

The formulae you need to memorise for factorisation on the Cambridge IGCSE 0580 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 algebraic expressions
When to use
Whenever you spot a subtraction of two terms that are both perfect squares.
Quadratic factorisation (monic)
$x^{2} + (p+q)x + pq = (x + p)(x + q)$
$p, q$
two numbers whose sum is the middle coefficient and whose product is the constant term
When to use
Use whenever the leading coefficient is $1$ — find the two numbers and write the brackets.
Perfect-square trinomial
$a^{2} \pm 2ab + b^{2} = (a \pm b)^{2}$
When to use
Recognise this when the first and last terms are perfect squares and the middle term is twice the product of their roots.

Key Definitions and Keywords — Factorisation

Definitions to memorise and the exact keywords mark schemes credit for factorisation answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Factorise
Examiner keyword
Rewrite an algebraic expression as a product of two or more factors.
Factorise fully
Examiner keyword
Factorise until no further common factor or factorisable structure remains.
Highest common factor (HCF) of algebraic terms
Examiner keyword
The largest expression that divides each given term exactly — found by taking the lower power of each shared variable and the HCF of the coefficients.
Example
HCF of $12x^{2}y$ and $18xy^{2}$ is $6xy$ .
Quadratic trinomial
An expression of the form $ax^{2} + bx + c$ with $a \neq 0$ .
Difference of two squares
Examiner keyword
An expression of the form $a^{2} - b^{2}$ , which always factors as $(a+b)(a-b)$ .

Common Mistakes and Misconceptions — Factorisation

The traps other students keep falling into on factorisation questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Stopping before the factorisation is fully done
0580/42 — examiner reports each series
Why it happens
Students take out a partial common factor and stop, e.g. $3(4x - 6y)$ instead of $6(2x - 3y)$ .
How to avoid it
After factoring, scan the bracket — if there's still a common factor, take it out.
✕Trying to factorise a sum of two squares as if it were a difference
Why it happens
$a^{2} + b^{2}$ does NOT factor over the reals; only $a^{2} - b^{2}$ does.
How to avoid it
Difference of two squares only — never sum of two squares.
✕Wrong signs in the quadratic brackets
Why it happens
Students find the right pair of numbers but assign the signs incorrectly.
How to avoid it
Always re-expand to check: $(x + p)(x + q) = x^{2} + (p+q)x + pq$ . Match coefficients.
✕Trying to factor $ax^{2} + bx + c$ (with $a \neq 1$ ) by inspection
0580 Extended — recurring
Why it happens
Inspection works for $a=1$ ; for $a \neq 1$ it usually doesn't.
How to avoid it
Use the splitting method: find two numbers whose product is $ac$ and sum is $b$ .

Factorisation — frequently asked questions

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

Factorisation

Short Notes - Factorisation

Detailed Study Notes

At a glance

What you’ll learn

Step 1 — pull out the highest common factor

Difference of two squares

Quadratic x2+bx+cx^{2} + bx + cx2+bx+c

Quadratic ax2+bx+cax^{2} + bx + cax2+bx+c with a≠1a \ne 1a=1

Four-term expressions: factorise by grouping

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Factorisation

1Take out the highest common factor

2Factorise a difference of two squares

3Factorise a quadratic trinomial

4Factorise by grouping

5Factorise 2x2+7x+32x^{2} + 7x + 32x2+7x+3

6Take out a numeric common factor

7Difference of two squares with a leading coefficient

8Factorise a quadratic with a negative constant

9Four-term grouping with a sign change

10Factorise a perfect-square trinomial

11Factorise using a surd factor

12Factorise using a common bracket

13Factorise an expression and use it to solve an equation

Difference of two squares

Quadratic factorisation (monic)

Perfect-square trinomial

Factorise

Factorise fully

Highest common factor (HCF) of algebraic terms

Quadratic trinomial

Difference of two squares

✕Stopping before the factorisation is fully done

✕Trying to factorise a sum of two squares as if it were a difference

✕Wrong signs in the quadratic brackets

✕Trying to factor ax2+bx+cax^{2} + bx + cax2+bx+c (with a≠1a \neq 1a=1) by inspection

Factorisation — frequently asked questions

Factorisation

Short Notes - Factorisation

Detailed Study Notes

At a glance

What you’ll learn

Step 1 — pull out the highest common factor

Difference of two squares

Quadratic x2+bx+cx^{2} + bx + cx2+bx+c

Quadratic ax2+bx+cax^{2} + bx + cax2+bx+c with a≠1a \ne 1a=1

Four-term expressions: factorise by grouping

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Factorisation

1Take out the highest common factor

2Factorise a difference of two squares

3Factorise a quadratic trinomial

4Factorise by grouping

5Factorise 2x2+7x+32x^{2} + 7x + 32x2+7x+3

6Take out a numeric common factor

7Difference of two squares with a leading coefficient

8Factorise a quadratic with a negative constant

9Four-term grouping with a sign change

10Factorise a perfect-square trinomial

11Factorise using a surd factor

12Factorise using a common bracket

13Factorise an expression and use it to solve an equation

Difference of two squares

Quadratic factorisation (monic)

Perfect-square trinomial

Factorise

Factorise fully

Highest common factor (HCF) of algebraic terms

Quadratic trinomial

Difference of two squares

✕Stopping before the factorisation is fully done

✕Trying to factorise a sum of two squares as if it were a difference

✕Wrong signs in the quadratic brackets

Quadratic $x^{2} + bx + c$

Quadratic $ax^{2} + bx + c$ with $a \ne 1$

5Factorise $2x^{2} + 7x + 3$

✕Trying to factor $ax^{2} + bx + c$ (with $a \neq 1$ ) by inspection

Quadratic $x^{2} + bx + c$

Quadratic $ax^{2} + bx + c$ with $a \ne 1$

5Factorise $2x^{2} + 7x + 3$

✕Trying to factor $ax^{2} + bx + c$ (with $a \neq 1$ ) by inspection