Factorise ab + 4a + 3b + 12.

a(b + 4) + 3(b + 4) = (b + 4)(a + 3)

Factorise 2x² + 5x - 3.

2x² + 6x - x - 3 = (2x - 1)(x + 3)

Factorise 6x² - 11x + 4.

6x² - 8x - 3x + 4 = (3x - 4)(2x - 1)

Free worksheet · Australian Curriculum

Factorising worksheets with answers

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Factorising at each year level, matched to the textbook chapters most Victorian schools use: a free printable worksheet with answers, sample questions, and an auto-marked test teachers can share.

Free Year 9 factorising worksheet

10 questions, then the answers with worked solutions. Printable PDF, no sign-up.

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Year 9 topics covered

Factorising algebraic expressions
Factorising the difference of two squares
Factorising by grouping in pairs
Factorising monic quadratic trinomials
Factorising trinomials of the form ax² + bx + c

Sample Year 9 factorising questions

Factorising algebraic expressions
Factorise fully:

(a) $15x - 25$ (b) $12ab + 18a$ (c) $-4m^2 - 10m$
Show the answer
Key idea: Take out the highest common factor (HCF) of the numbers and the pronumerals. If the first term is negative, take out a negative factor.
1. (a) HCF $= 5$ : $15x - 25 = 5(3x - 5)$
2. (b) HCF $= 6a$ : $12ab + 18a = 6a(2b + 3)$
3. (c) HCF $= -2m$ : $-4m^2 - 10m = -2m(2m + 5)$
Answer: (a) $5(3x - 5)$ (b) $6a(2b + 3)$ (c) $-2m(2m + 5)$
Factorising the difference of two squares
Factorise:

(a) $x^2 - 36$ (b) $49 - a^2$ (c) $4m^2 - 25$
Show the answer
Key idea: A difference of two squares factorises as $a^2 - b^2 = (a - b)(a + b)$ .
1. (a) $x^2 - 6^2 = (x - 6)(x + 6)$
2. (b) $7^2 - a^2 = (7 - a)(7 + a)$
3. (c) $(2m)^2 - 5^2 = (2m - 5)(2m + 5)$
Answer: (a) $(x - 6)(x + 6)$ (b) $(7 - a)(7 + a)$ (c) $(2m - 5)(2m + 5)$
Factorising by grouping in pairs
Factorise $x^2 + 3x + 2x + 6$ by grouping in pairs.
Show the answer
Key idea: Group the four terms into two pairs, take out a common factor from each pair, then take out the common bracket.
1. $x(x + 3) + 2(x + 3)$
2. $(x + 3)(x + 2)$
Answer: $(x + 3)(x + 2)$
Factorising monic quadratic trinomials
Factorise $x^2 - x - 12$ .
Show the answer
Key idea: Find two integers that multiply to the constant term and add to the coefficient of $x$ . Write the expression as a product of two binomial factors.
1. Find two numbers that multiply to $-12$ and add to $-1$ .
2. Test factor pairs of $-12$ : $3 \times (-4) = -12$ and $3 + (-4) = -1$ . ✓
3. Write the factorised form: $(x + 3)(x - 4)$
4. Check by expanding: $(x + 3)(x - 4) = x^2 - 4x + 3x - 12 = x^2 - x - 12$ ✓
Answer: $(x + 3)(x - 4)$

Free Year 10 factorising worksheet

10 questions, then the answers with worked solutions. Printable PDF, no sign-up.

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Teachers: share one link and every student is marked straight away, with worked solutions.

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Printable practice exam

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Year 10 topics covered

Factorising expressions
Factorising monic quadratic trinomials

Sample Year 10 factorising questions

Factorising expressions
Factorise $6x^3 + 9x^2 - 3x$ .
Show the answer
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: $3x(2x^2 + 3x - 1)$
Factorising monic quadratic trinomials
Factorise $x^2 + 7x + 12$ .
Show the answer
Key idea: Find two integers that multiply to $12$ and add to $7$ .
1. List factor pairs of $12$ : $(1, 12),\ (2, 6),\ (3, 4)$ .
2. Check which pair adds to $7$ : $3 + 4 = 7$ . ✓
3. Write the factorised form: $(x + 3)(x + 4)$ .
4. Verify by expanding: $(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ . ✓
Answer: $(x + 3)(x + 4)$
Factorising expressions
Factorise $-12a^3b^2 + 8a^2b^3 - 4a^2b$ .
Show the answer
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: $-4a^2b(3ab - 2b^2 + 1)$
Factorising monic quadratic trinomials
Factorise each expression.

(a) $x^2 - 49$

(b) $x^2 - 10x + 25$
Show the answer
Key idea: Recognise special quadratic forms: $a^2 - b^2 = (a-b)(a+b)$ and $a^2 - 2ab + b^2 = (a - b)^2$ .

Part (a)
1. Write $x^2 - 49$ as $x^2 - 7^2$ .
2. Apply the difference of two squares rule: $(x - 7)(x + 7)$ .
Part (b)
1. Check: $\sqrt{25} = 5$ and the middle term $-10x = -2 \times x \times 5$ . ✓
2. This is a perfect-square trinomial: $(x - 5)^2$ .
3. Verify: $(x-5)^2 = x^2 - 10x + 25$ . ✓
Answer: (a) $(x - 7)(x + 7)$ (b) $(x - 5)^2$

Free Year 10A factorising worksheet

10 questions, then the answers with worked solutions. Printable PDF, no sign-up.

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Auto-marked test for your class

Teachers: share one link and every student is marked straight away, with worked solutions.

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Printable practice exam

Mix in harder questions, with marks and a suggested time.

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Year 10A topics covered

Factorising expressions
Factorising monic quadratic trinomials
Factorising non-monic quadratic trinomials10A

Sample Year 10A factorising questions

Factorising expressions
Factorise $6x^3 + 9x^2 - 3x$ .
Show the answer
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: $3x(2x^2 + 3x - 1)$
Factorising monic quadratic trinomials
Factorise $x^2 + 7x + 12$ .
Show the answer
Key idea: Find two integers that multiply to $12$ and add to $7$ .
1. List factor pairs of $12$ : $(1, 12),\ (2, 6),\ (3, 4)$ .
2. Check which pair adds to $7$ : $3 + 4 = 7$ . ✓
3. Write the factorised form: $(x + 3)(x + 4)$ .
4. Verify by expanding: $(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ . ✓
Answer: $(x + 3)(x + 4)$
Factorising non-monic quadratic trinomials
Factorise $6x^2 + 11x + 3$ .
Show the answer
Key idea: For a non-monic quadratic $ax^2 + bx + c$ , find two numbers that multiply to $ac$ and add to $b$ , then split the middle term and factorise by grouping.
1. Identify $a = 6$ , $b = 11$ , $c = 3$ , so $ac = 6 \times 3 = 18$ .
2. Find two numbers that multiply to $18$ and add to $11$ : these are $2$ and $9$ .
3. Split the middle term: $6x^2 + 2x + 9x + 3$ .
4. Factorise by grouping in pairs: $2x(3x + 1) + 3(3x + 1)$ .
5. Factor out the common bracket: $(3x + 1)(2x + 3)$ .
Answer: $(3x + 1)(2x + 3)$
Factorising expressions
Factorise $-12a^3b^2 + 8a^2b^3 - 4a^2b$ .
Show the answer
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: $-4a^2b(3ab - 2b^2 + 1)$

Topic guide

Factorising, explained

Always look for a common factor first, then count the terms to choose the method.

1. Take out the highest common factor

Find the HCF of every term, write it outside the bracket, and divide each term by it:

$6x + 15 = 3(2x + 5) 8a^2 - 12a = 4a(2a - 3)$

Expand your answer to check it gives back the original.

2. The difference of two squares

Two perfect squares with a minus between them factorise as

$a^2 - b^2 = (a - b)(a + b)$

$x^2 - 49 = (x - 7)(x + 7) 9y^2 - 16 = (3y - 4)(3y + 4)$

Take out a common factor first if there is one: $2x^2 - 50 = 2(x - 5)(x + 5)$ .

3. Monic quadratic trinomials

For $x^2 + bx + c$ , find two numbers that multiply to $c$ and add to $b$ . For $x^2 + 7x + 12$ , the numbers are $3$ and $4$ :

$x^2 + 7x + 12 = (x + 3)(x + 4)$

Watch the signs: $x^2 - 2x - 15 = (x - 5)(x + 3)$ , because $-5 \times 3 = -15$ and $-5 + 3 = -2$ .

4. Non-monic quadratic trinomials

For $ax^2 + bx + c$ , find two numbers that multiply to $ac$ and add to $b$ , split the middle term, then group. For $2x^2 + 7x + 3$ , $ac = 6$ , and $6$ and $1$ add to $7$ :

$2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$

5. Grouping in pairs

With four terms, factorise each pair, then take out the common bracket:

$xy + 5x + 2y + 10 = x(y + 5) + 2(y + 5) = (y + 5)(x + 2)$

If the brackets don't match, try pairing the terms a different way.

6. Putting it together

HCF first, then count the terms: two terms suggests the difference of two squares, three terms a trinomial, four terms grouping.

$3x^2 - 12x - 36 = 3(x^2 - 4x - 12) = 3(x - 6)(x + 2)$

An expression is fully factorised when no bracket can be factorised further.

Quick check: 10 questions

Try each one, then open the answer.

Factorise $5x + 20$ .
Answer
$5(x + 4)$
Factorise $12a^2 - 18a$ .
Answer
$6a(2a - 3)$
Factorise $x^2 - 36$ .
Answer
$(x - 6)(x + 6)$
Factorise $x^2 + 9x + 20$ .
Answer
$(x + 4)(x + 5)$
Factorise $x^2 - x - 12$ .
Answer
$(x - 4)(x + 3)$
Factorise $25m^2 - 4$ .
Answer
$(5m - 2)(5m + 2)$
Factorise $3x^2 - 27$ fully.
Answer
$3(x^2 - 9) = 3(x - 3)(x + 3)$
Factorise $ab + 4a + 3b + 12$ .
Answer
$a(b + 4) + 3(b + 4) = (b + 4)(a + 3)$
Factorise $2x^2 + 5x - 3$ .
Answer
$2x^2 + 6x - x - 3 = (2x - 1)(x + 3)$
Factorise $6x^2 - 11x + 4$ .
Answer
$6x^2 - 8x - 3x + 4 = (3x - 4)(2x - 1)$

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These topic pages cover this skill, with worked examples, common mistakes and a question to try.

Year 10 worksheetQuadratic factorising Year 10 worksheetFactorising by common factor Year 9 worksheetQuadratics Year 10A worksheetFactorising non-monic quadratics Year 8 worksheetAlgebra

Questions

What does factorising mean?

Factorising means writing an expression as a product of its factors, usually with brackets. It is the reverse of expanding, so expanding your answer is a quick way to check it.

What should you look for first when factorising?

Always take out the highest common factor first. Then count the terms: two terms may be a difference of two squares, three terms a quadratic trinomial, and four terms can often be grouped in pairs.

Do the factorising worksheets come with answers?

Yes. Every question on a generated worksheet has a fully worked solution, and you can choose a year level to get a free printable PDF.

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Factorising, explained

1. Take out the highest common factor

2. The difference of two squares

3. Monic quadratic trinomials

4. Non-monic quadratic trinomials

5. Grouping in pairs

6. Putting it together

Quick check: 10 questions

Generate a free maths worksheet

Practice built from mistakes.

Topic worksheets with worked examples

Questions

What does factorising mean?

What should you look for first when factorising?

Do the factorising worksheets come with answers?

More skill worksheets

Better maths marks start with the next wrong answer.