Factorising Common Factors
Expanding puts brackets in; factorising pulls them out. Learn to spot what divides every term, extract it, and check your answer by expanding back.
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Before you read on — factorise $8x + 12$. What number divides both terms? What is left inside the brackets? Try it, then check your reasoning as you go.
Factorising is the reverse of expanding. Where expanding multiplies a bracket out, factorising searches for what every term shares, pulls it out front, and leaves the rest in brackets.
An expansion turns $3(x + 4)$ into $3x + 12$. A factorisation turns $3x + 12$ back into $3(x + 4)$. The move is reversible — and you verify it by expanding back.
Know
- What factorising means and why it is the reverse of expanding
- How to find the HCF of numbers and variables
- The grouping method for four-term expressions
Understand
- Why the HCF uses the smallest power of each variable
- Why every term must be divided by the HCF
- How grouping turns four terms into a common binomial
Can Do
- Find the HCF of any set of algebraic terms
- Factorise by extracting the HCF
- Factorise by grouping
- Verify by expanding back
Wrong: "$6x + 9 = 3(2x + 9)$" — the 9 was not divided by 3.
Right: $6x + 9 = 3(2x + 3)$. Every term must be divided by the HCF.
Wrong: "$10a^2b + 15ab = 5ab(2a^2 + 3)$" — the $a^2$ was not divided by $a$.
Right: $10a^2b + 15ab = 5ab(2a + 3)$. Divide variables as well as numbers.
Before you can factorise, you must find the Highest Common Factor — the biggest thing that divides every term. Check numbers first, then variables, then powers.
For numbers, find the HCF of the coefficients. For variables, use the smallest power that appears in every term. $x^2$ and $x$ share $x$ (not $x^2$).
and $9xy^3$ is $3xy$
Example: Find the HCF of $6x^2y$ and $9xy^3$.
- Numbers: HCF of 6 and 9 is $3$
- $x$: smallest power is $x$ (i.e. $x^1$)
- $y$: smallest power is $y$ (i.e. $y^1$)
So the HCF is $3xy$.
Once you know the HCF, divide every term by it. Write the HCF outside the brackets and the quotients inside. Nothing should remain un-divided.
The HCF is the multiplier that rebuilds the original when distributed. Divide each term by the HCF to find what sits inside the brackets. Check by expanding.
= 4(2x + 3)
Step-by-step: factorise $10a^2b + 15ab^2$.
$$\text{HCF} = 5ab$$
$$\frac{10a^2b}{5ab} = 2a \qquad \frac{15ab^2}{5ab} = 3b$$
$$10a^2b + 15ab^2 = 5ab(2a + 3b)$$
The fastest way to catch a mistake is to expand your answer. If you don't get back to the original expression, something went wrong. This one habit saves more marks than any other.
Think of factorising and expanding as round-trip tickets. You should arrive back where you started. If $3(2x + 3)$ expands to $6x + 9$, you know the factorisation is correct.
= 6x + 9 check
When an expression has four terms and no single HCF, split it into two pairs. Factorise each pair separately. If the same bracket appears in both, pull it out as a common factor.
Group $(ax + ay)$ and $(bx + by)$. Factorise each: $a(x + y)$ and $b(x + y)$. The bracket $(x + y)$ is common — extract it to get $(a + b)(x + y)$.
= (a+b)(x+y)
Example: factorise $2x + 6 + xy + 3y$.
$$\text{Group: } (2x + 6) + (xy + 3y)$$
$$= 2(x + 3) + y(x + 3)$$
$$= (x + 3)(2 + y)$$
Watch Me Solve It · 3 examples
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1Find the HCF of the coefficientsHCF of 8 and 12 is $4$Check: 4 divides 8 and 4 divides 12. No larger number does both.
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2Divide each term by the HCF$\dfrac{8x}{4} = 2x$ and $\dfrac{12}{4} = 3$Both terms are divided by 4. The variable $x$ has no partner in 12, so it stays with the first term.
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3Write in factorised form$8x + 12 = 4(2x + 3)$4 outside, $2x + 3$ inside. No further common factors exist.
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4Check by expanding$4(2x + 3) = 8x + 12$ ✓We get back to the original — the factorisation is correct.
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1Find the HCF — numbersHCF of 10 and 15 is $5$
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2Find the HCF — variables$a^2b$ and $ab^2$ share $ab$Smallest power of $a$ is $a^1$. Smallest power of $b$ is $b^1$.
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3Combine and divideOverall HCF = $5ab$$\dfrac{10a^2b}{5ab} = 2a$ and $\dfrac{15ab^2}{5ab} = 3b$
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4Write and check$10a^2b + 15ab^2 = 5ab(2a + 3b)$Expand: $5ab \times 2a = 10a^2b$ and $5ab \times 3b = 15ab^2$ ✓
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1Group into pairs$(2x + 6) + (xy + 3y)$First two share a factor of 2. Last two share a factor of $y$.
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2Factorise each pair$2(x + 3) + y(x + 3)$Both pairs produce the identical bracket $(x + 3)$.
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3Extract the common bracket$(x + 3)(2 + y)$$(x + 3)$ is common to both terms. The leftovers ($2$ and $y$) form the second bracket.
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4Check by expanding$(x + 3)(2 + y) = 2x + xy + 6 + 3y$ ✓Rearrange to match the original: $2x + 6 + xy + 3y$.
The HCF
- Highest common factor of coefficients
- Smallest power of each variable
- Combine: number × variables
Extraction
- Divide every term by HCF
- Write HCF outside brackets
- Quotients go inside
Check
- Expand your answer
- Must match original
- Catches division errors
Grouping
- Split 4 terms into 2 pairs
- Factorise each pair
- Extract common bracket
How are you completing this lesson?
Brain Trainer · 4 problems
Four problems mixing basic HCF, variables and grouping. Work each one, then reveal the answer to check.
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1 Factorise $14m + 21n$.
HCF of 14 and 21 is 7$= 7(2m + 3n)$ -
2 Factorise $9x^2 - 12x$.
HCF of 9 and 12 is 3. HCF of $x^2$ and $x$ is $x$$= 3x(3x - 4)$ -
3 Factorise $6a^2b + 9ab^2$.
HCF of 6 and 9 is 3. HCF of $a^2b$ and $ab^2$ is $ab$$= 3ab(2a + 3b)$ -
4 Factorise by grouping: $3x + 3y + ax + ay$.
Group: $(3x + 3y) + (ax + ay) = 3(x + y) + a(x + y)$$= (x + y)(3 + a)$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Fully factorise each of the following expressions.
(a) $14m + 21n$ (1 mark)
(b) $9x^2 - 12x$ (1 mark)
Q7. Factorise by grouping: $2x + 6 + xy + 3y$. Show all steps clearly.
Q8. A rectangular garden has an area of $(8x + 12)$ square metres. The width of the garden is 4 metres.
(a) Factorise the expression for the area. (1 mark)
(b) Hence find an expression for the length of the garden in factorised form. (2 marks)
Quick Check
1. B — HCF of 12 and 18 is 6. Smallest power of $x$ is $x^1$ and of $y$ is $y^1$. So HCF is $6xy$.
2. B — HCF is 3. $6x \div 3 = 2x$ and $9 \div 3 = 3$. So $3(2x + 3)$.
3. A — HCF is $5ab$. $10a^2b \div 5ab = 2a$ and $15ab \div 5ab = 3$. So $5ab(2a + 3)$.
4. A — Group: $a(x+y) + b(x+y) = (a+b)(x+y)$.
5. C — $2(3x + 6)$ is not fully factorised because $3x + 6$ still has a common factor of 3. Correct: $6(x + 2)$.
Show Your Working Model Answers
Q6 (2 marks): (a) $7(2m + 3n)$ [1]. (b) HCF is $3x$, so $3x(3x - 4)$ [1].
Q7 (2 marks): $(2x + 6) + (xy + 3y)$ [1] $= 2(x + 3) + y(x + 3) = (x + 3)(2 + y)$ [1].
Q8 (3 marks): (a) $4(2x + 3)$ [1]. (b) Area = width $\times$ length, so $4(2x + 3) = 4 \times \text{length}$. Length = $2x + 3$ metres [2].
Factorise in Two Stages
Factorise $6x^2 + 12x + 9x + 18$ completely. Show all steps: group first, factorise each pair, extract the common bracket, then check if any further factorising is possible inside the brackets.
Reveal solution
Group: $(6x^2 + 12x) + (9x + 18) = 6x(x + 2) + 9(x + 2)$.
Extract: $(x + 2)(6x + 9)$.
Further factorise: $6x + 9 = 3(2x + 3)$.
Fully factorised: $3(x + 2)(2x + 3)$.
HCF
Highest common factor of numbers and variables
Variables
Use the smallest power of each variable
Divide
Divide every term by the HCF
Check
Expand to verify your answer
Grouping
Group in pairs, factorise each, then common bracket
Fully Factorised
No common factor remains inside brackets
Interactive: Factor Extractor
Practise finding the HCF and factorising expressions with this interactive tool. Adjust the terms and watch the factor extraction step by step.