What does factorise mean?

Writing an expression as a product: something multiplied by something. It's expanding in reverse. Expanding 3(x + 4) gives 3x + 12, so factorising 3x + 12 gives 3(x + 4).

What should I always do first?

Look for a common factor (the HCF) and take it out. Then count the terms: two terms might be a difference of two squares, three terms is usually a quadratic trinomial, and four terms usually means grouping in pairs.

How do I factorise x² + bx + c?

Find two numbers that multiply to c and add to b. If they're p and q, the answer is (x + p)(x + q). For x² + 7x + 12, the numbers are 3 and 4, so it factorises to (x + 3)(x + 4).

What if a quadratic won't factorise?

Not every quadratic factorises using whole numbers. If b² − 4ac isn't a perfect square, it doesn't. To solve the equation anyway, use the quadratic formula or complete the square.

What year level learns factorising?

In the Victorian Curriculum 2.0, taking out common factors starts in Year 8, factorising monic quadratics like x² + 5x + 6 is Year 9, and Year 10 adds more quadratic factorising. Non-monic quadratics like 2x² + 7x + 3 are usually Year 10A. Schools sequence topics differently, so check your course.

How to Factorise: Methods, Examples and Factorising Calculator (Years 8 to 10)

What factorising means

Expanding turns a product into a sum. Factorising goes the other way.

3(x + 4) \;\xrightarrow{\text{expand}}\; 3x + 12

3x + 12 \;\xrightarrow{\text{factorise}}\; 3(x + 4)

That means you can always check a factorising answer by expanding it. If you get back to the question, it's right.

Taking out a common factor (HCF)

Find the highest common factor of every term: the biggest number and the most of each letter that goes into all of them. Write it outside the bracket, and divide each term by it to get what goes inside.

Worked examples

$6x + 9$ : the HCF is $3$ .

6x + 9 = 3(2x + 3)

$12x^2y - 8xy$ : the HCF is $4xy$ (4 goes into 12 and 8, and both terms have an $x$ and a $y$ ).

12x^2y - 8xy = 4xy(3x - 2)

$-5x - 15$ : when the first term is negative, take out a negative factor.

-5x - 15 = -5(x + 3)

If you take out a factor that isn't the highest, like $2(6x^2y - 4xy)$ , the answer isn't fully factorised. Look inside the bracket again.

Factorising quadratics x² + bx + c

x^2 + bx + c = (x + p)(x + q) \quad \text{where } p \times q = c \text{ and } p + q = b

Find two numbers that multiply to $c$ and add to $b$ . List the factor pairs of $c$ and check which pair adds to $b$ .

Worked example

Factorise $x^2 + 7x + 12$ .

Factor pairs of 12: $1 \times 12$ , $2 \times 6$ , $3 \times 4$ . Only $3 + 4 = 7$ .

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

Worked example with negatives

Factorise $x^2 - 2x - 15$ .

You need two numbers that multiply to $-15$ and add to $-2$ : $-5$ and $3$ .

x^2 - 2x - 15 = (x - 5)(x + 3)

Factorise $x^2 - 9x + 20$ : $-4 \times -5 = 20$ and $-4 + (-5) = -9$ .

x^2 - 9x + 20 = (x - 4)(x - 5)

Sign shortcut:

$c$ positive, $b$ positive: both numbers are positive.
$c$ positive, $b$ negative: both numbers are negative.
$c$ negative: one is positive and one is negative. The bigger one (ignoring signs) has the same sign as $b$ .

Factorising calculator (with working)

Enter a, b and c for any quadratic ax² + bx + c with whole-number coefficients. It takes out the common factor first and shows how to find the two numbers.

Difference of two squares

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

Look for two terms, both perfect squares, with a minus between them.

Worked examples

$x^2 - 25 = x^2 - 5^2$

x^2 - 25 = (x - 5)(x + 5)

$4x^2 - 9 = (2x)^2 - 3^2$

4x^2 - 9 = (2x - 3)(2x + 3)

$2x^2 - 50$ : take out the common factor first.

2x^2 - 50 = 2(x^2 - 25) = 2(x - 5)(x + 5)

A sum of two squares, like $x^2 + 25$ , doesn't factorise this way.

Perfect squares

a^2 + 2ab + b^2 = (a + b)^2 \qquad a^2 - 2ab + b^2 = (a - b)^2

If the first and last terms are perfect squares and the middle term is twice their square roots multiplied, it's a perfect square.

Worked examples

$x^2 + 6x + 9$ : $x^2$ and $9 = 3^2$ are squares, and $2 \times x \times 3 = 6x$ .

x^2 + 6x + 9 = (x + 3)^2

$9x^2 - 12x + 4$ : $9x^2 = (3x)^2$ , $4 = 2^2$ , and $2 \times 3x \times 2 = 12x$ .

9x^2 - 12x + 4 = (3x - 2)^2

Grouping in pairs (four terms)

With four terms, split them into two pairs, take a common factor out of each pair, then take out the bracket they share.

Worked examples

$x^3 + 3x^2 + 2x + 6$

= x^2(x + 3) + 2(x + 3)

= (x + 3)(x^2 + 2)

$xy - 2y + 3x - 6$

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

= (x - 2)(y + 3)

If the two brackets don't match, try swapping the middle two terms and grouping again.

Non-monic quadratics ax² + bx + c (Year 10A)

When the $x^2$ term has a number in front (and there's no common factor), split the middle term:

Multiply $a \times c$ .
Find two numbers that multiply to $ac$ and add to $b$ .
Use them to split $bx$ into two terms.
Factorise by grouping in pairs.

Worked example

Factorise $2x^2 + 7x + 3$ .

$ac = 2 \times 3 = 6$ . Two numbers that multiply to 6 and add to 7: $6$ and $1$ .

2x^2 + 6x + x + 3

= 2x(x + 3) + 1(x + 3)

= (x + 3)(2x + 1)

Worked example with negatives

Factorise $6x^2 - x - 2$ .

$ac = 6 \times (-2) = -12$ . Two numbers that multiply to $-12$ and add to $-1$ : $-4$ and $3$ .

6x^2 - 4x + 3x - 2

= 2x(3x - 2) + 1(3x - 2)

= (3x - 2)(2x + 1)

Which method should I use?

Common factor first, every time. $3x^2 - 12x - 36 = 3(x^2 - 4x - 12)$ .
Count the terms in what's left:
- Two terms with a minus: difference of two squares.
- Three terms: a quadratic trinomial. Check for a perfect square, then find the two numbers.
- Four terms: grouping in pairs.
Check each bracket to see if it factorises again.
Expand to check.

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

Practice questions with answers

Try each one before you open the answer.

Factorise $8x + 12$ .
Show the answer
The HCF is $4$ : $4(2x + 3)$
Factorise $15a^2b + 10ab^2$ .
Show the answer
The HCF is $5ab$ : $5ab(3a + 2b)$
Factorise $x^2 + 9x + 20$ .
Show the answer
$4 \times 5 = 20$ and $4 + 5 = 9$ , so $(x + 4)(x + 5)$
Factorise $x^2 - x - 12$ .
Show the answer
$-4 \times 3 = -12$ and $-4 + 3 = -1$ , so $(x - 4)(x + 3)$
Factorise $x^2 - 49$ .
Show the answer
Difference of two squares: $(x - 7)(x + 7)$
Factorise $2x^2 - 8x - 24$ fully.
Show the answer
Take out $2$ : $2(x^2 - 4x - 12)$ . Then $-6 \times 2 = -12$ and $-6 + 2 = -4$ , so $2(x - 6)(x + 2)$
Factorise $ab + 3a + 2b + 6$ .
Show the answer
Group in pairs: $a(b + 3) + 2(b + 3) = (b + 3)(a + 2)$
Factorise $3x^2 + 10x + 8$ (Year 10A).
Show the answer
$ac = 24$ . $6 \times 4 = 24$ and $6 + 4 = 10$ . Split: $3x^2 + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2) = (x + 2)(3x + 4)$

Factorising by year level

Following the Victorian Curriculum 2.0. Schools sequence topics differently, so check your course.

Year 8

Algebra

Factorising by taking out a numerical or algebraic common factor.

Worksheets and topic test · Free PDF with answers

Year 9

Algebra

Factorising monic quadratics x² + bx + c.

Worksheets and topic test · Free PDF with answers

Year 10

Quadratic factorising

Quadratic factorising, including difference of two squares and perfect squares.

Worksheets and topic test · Free PDF with answers

Year 10A

Factorising non-monic quadratics

Non-monic quadratics ax² + bx + c by splitting the middle term.

Worksheets and topic test · Free PDF with answers

How to factorise

What factorising means

Taking out a common factor (HCF)

Factorising quadratics x² + bx + c

Factorising calculator (with working)

Difference of two squares

Perfect squares

Grouping in pairs (four terms)

Non-monic quadratics ax² + bx + c (Year 10A)

Which method should I use?

Practice questions with answers

Factorising by year level

Factorising: common questions

More maths help, Years 7 to 10

Got a factorising question wrong in homework?