Year 10 - Factorising by common factor - Free practice

Year 10 Factorising by Taking Out a Common Factor Worksheets & Practice

Spot the common factor, write the brackets, check your answer. Worked solutions every time, free.

Quick answer

How do you factorise an expression by taking out a common factor?

Find the highest common factor (HCF) of all terms, then divide each term by it and write the result in brackets. For example, 6x² + 9x has HCF 3x, so it factorises to 3x(2x + 3). Check by expanding: the brackets should reproduce the original expression.

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Worked examples

Year 10 factorising by common factor: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Factorising with a numerical and single-variable common factor
Factorise $6x^3 + 9x^2 - 3x$ .
Show worked solution
Key idea: Find the HCF of all terms, write it outside the brackets, then divide each term by the HCF to find what goes inside.
1. Find the HCF of the coefficients $6$ , $9$ and $3$ : the HCF is $3$ .
2. Find the HCF of $x^3$ , $x^2$ and $x$ : the lowest power is $x^1 = x$ .
3. So the overall HCF is $3x$ .
4. Divide each term by $3x$ : $\frac{6x^3}{3x} = 2x^2, \frac{9x^2}{3x} = 3x, \frac{-3x}{3x} = -1$
5. Write the factorised form: $6x^3 + 9x^2 - 3x = 3x(2x^2 + 3x - 1)$
Check: $3x(2x^2 + 3x - 1) = 6x^3 + 9x^2 - 3x$ ✓

Answer: $3x(2x^2 + 3x - 1)$
Example 2 · Factorising with a multi-variable algebraic common factor and a negative
Factorise $-12a^3b^2 + 8a^2b^3 - 4a^2b$ .
Show worked solution
Key idea: When the leading term is negative, take out a negative common factor so the first term inside the brackets is positive. Find the full HCF including all variables.
1. Find the HCF of the coefficients $12$ , $8$ and $4$ : the HCF is $4$ .
2. Include the negative sign: take out $-4$ .
3. Find the HCF of the variable parts: lowest power of $a$ is $a^2$ ; lowest power of $b$ is $b^1 = b$ . So the variable HCF is $a^2b$ .
4. Overall factor to take out: $-4a^2b$ .
5. Divide each term by $-4a^2b$ : $\frac{-12a^3b^2}{-4a^2b} = 3ab, \frac{8a^2b^3}{-4a^2b} = -2b^2, \frac{-4a^2b}{-4a^2b} = 1$
6. Write the factorised form: $-12a^3b^2 + 8a^2b^3 - 4a^2b = -4a^2b(3ab - 2b^2 + 1)$
Check: $-4a^2b(3ab - 2b^2 + 1) = -12a^3b^2 + 8a^2b^3 - 4a^2b$ ✓

Answer: $-4a^2b(3ab - 2b^2 + 1)$
Example 3 · Factorising with a binomial common factor
A rectangular garden bed has an area that can be expressed as $5x(x+3) - 2(x+3)$ square metres, where $x$ is a positive number. Write the area as a product of two factors, and find the area when $x = 4$ .
Show worked solution
Key idea: When the same bracket appears in every term, the whole bracket is a common factor and can be taken out just like a single term.
1. Identify the common binomial factor: $(x + 3)$ appears in both terms.
2. Take out $(x + 3)$ : $5x(x+3) - 2(x+3) = (x+3)(5x-2)$
3. Check by expanding: $(x+3)(5x-2) = 5x^2 - 2x + 15x - 6 = 5x^2 + 13x - 6$ , and $5x(x+3) - 2(x+3) = 5x^2 + 15x - 2x - 6 = 5x^2 + 13x - 6$ ✓
4. Substitute $x = 4$ : $\text{Area} = (4+3)(5 \times 4 - 2) = 7 \times 18 = 126 \text{ square metres}$
Answer: $(x+3)(5x-2)$ ; the area when $x = 4$ is $126$ square metres.

Common mistakes

Where students usually lose marks

Only taking out a partial common factor

HCF of $6x^3 + 9x^2 - 3x$ taken as $3$ instead of $3x$ , giving $3(2x^3 + 3x^2 - x)$ .

Always look for the HCF of both the coefficients and the variable parts. Check that nothing inside the brackets still shares a factor with the one you took out.

Sign errors when taking out a negative factor

$-4a^2b(3ab - 2b^2 + 1)$ written incorrectly as $-4a^2b(3ab + 2b^2 - 1)$ after dividing $8a^2b^3$ by $-4a^2b$ .

When you divide a positive term by a negative factor the result is negative, and vice versa. After factorising, always expand your answer to check every sign matches the original expression.

Forgetting the +1 (or -1) when a whole term is taken out

$3x(2x^2 + 3x)$ written for $6x^3 + 9x^2 - 3x$ , dropping the $-3x$ term entirely.

Every term in the original expression must appear inside the brackets. When a term equals the HCF exactly, dividing it by the HCF gives $1$ (or $-1$ ), which must be written inside the brackets.

Treating a binomial factor as two separate factors

$5x(x+3) - 2(x+3)$ rewritten as $(5x-2)(x)(3)$ by splitting the bracket $(x+3)$ apart.

Treat the entire bracket $(x+3)$ as a single unit, just like a pronumeral. Take it out as one factor: $(x+3)(5x-2)$ .

Try one yourself

Your turn

Factorise $12a^3 - 8a^2$ by taking out the highest common factor. Enter your answer as an expression.

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About this worksheet

Why we built it

Factorising by taking out a common factor is the essential first step in almost every algebraic factorising problem in Year 10. These worksheets build the skill from numerical common factors through to algebraic ones, including expressions where the common factor is itself a binomial, with full worked solutions so students can see exactly where their thinking needs to shift.

What's covered

Sub-skills your student will practise

✓Identifying the highest common factor (HCF) of terms
✓Factorising expressions with numerical common factors
✓Factorising with single-variable algebraic common factors
✓Factorising with multi-variable algebraic common factors
✓Taking out a negative common factor
✓Factorising when the common factor is a binomial expression

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Frequently asked

Questions parents ask about factorising by common factor

What does 'taking out a common factor' actually mean?

It means rewriting an expression as a product by dividing every term by the highest common factor and placing that factor outside a pair of brackets. For example, 6x² + 9x becomes 3x(2x + 3).

How do students find the highest common factor of algebraic terms?

Find the HCF of the coefficients first, then include any variables that appear in every term, using the lowest power of each. For 4x³ and 6x², the HCF is 2x².

Why do teachers ask students to 'check by expanding'?

Expanding the brackets after factorising multiplies everything back out, so the result should match the original expression exactly. It is a quick, reliable way to catch errors in the factorised form.

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