Simple Factorising

Taking out the highest common factor of numbers and letters, and factorising a square minus a square.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 102 real past-paper questions below

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What Factorising Means

Expanding takes the bracket away: $6 (2 y - 3) = 12 y - 18$ . Factorising puts it back: you start with $12 y - 18$ and find $6 (2 y - 3)$ .

Example

Think of 12. Writing it as $3 \times 4$ is factorising a number. Here you write an expression as something times a bracket.

Rule

Find what every term shares. Take it outside the bracket. What is left goes inside.

Key idea

You can always check: expand your answer and you should get the question back.

Take Out the Biggest Number

The 1-mark questions look like $9 y - 12$ . Find the biggest number that divides every term.

Worked example

1
$12 y - 18$ — The question. Look for the biggest number that goes into both 12 and 18.
2
$12 y = 6 \times 2 y, 18 = 6 \times 3$ — 6 is the biggest number that goes into 12 and 18. 2 and 3 also go in, but they are smaller.
3
$= 6 (2 y - 3)$ — Write the 6 once, outside the bracket. Inside goes what is left of each term, 2y and 3, with the minus kept between them.
4
$6 \times 2 y - 6 \times 3 = 12 y - 18 ✓$ — Expand to check. You get the question back, so the answer is right.

Common mistake

$3 (4 y - 6)$ is not finished: 4y and 6 still share 2. Use the biggest common factor.

Factorise Fully: Numbers and Letters

The 2-mark questions say fully. Take out the biggest number and every letter the terms share.

Rule

For each letter, take the lowest power that appears in every term. $a^{3}$ and $a^{2}$ share $a^{2}$ .

Worked example

1
$18 a^{3} b^{2} - 12 a^{2} b^{5}$ — The question. Deal with the numbers first, then each letter, one at a time.
2
$18 = 6 \times 3, 12 = 6 \times 2$ — 6 is the biggest number that goes into 18 and 12.
3
$a^{3}, a^{2} \to a^{2}$ — Both terms have at least $a^{2}$ , so a² comes out. Always take the lowest power.
4
$b^{2}, b^{5} \to b^{2}$ — Both terms have at least $b^{2}$ , so b² comes out too.
5
$common factor = 6 a^{2} b^{2}$ — Put the pieces together. This goes outside the bracket.
6
$18 a^{3} b^{2} \div 6 a^{2} b^{2} = 3 a$ — Divide the first term by it: 18 ÷ 6 = 3, $a^{3} \div a^{2} = a$ , and $b^{2} \div b^{2} = 1$ .
7
$12 a^{2} b^{5} \div 6 a^{2} b^{2} = 2 b^{3}$ — Divide the second term: 12 ÷ 6 = 2, $a^{2} \div a^{2} = 1$ , and $b^{5} \div b^{2} = b^{3}$ .
8
$= 6 a^{2} b^{2} (3 a - 2 b^{3})$ — Common factor outside, the two answers inside. Keep the minus from the question.

Common mistake

Taking out only the number: $6 (3 a^{3} b^{2} - 2 a^{2} b^{5})$ . The terms still share $a^{2} b^{2}$ , so it is not fully factorised.

Key idea

Show what you take out. In a 2-mark “fully” question, a correct but unfinished answer such as $2 p (6 p - 9)$ for $12 p^{2} - 18 p$ usually still earns 1 mark. Then check the bracket: 6p and 9 still share 3, so the full answer is $6 p (2 p - 3)$ . If a question asks why someone lost a mark, that is the reason to give.

When a Term Is the Whole Factor

Sometimes one term is the whole common factor, like the 4x in $4 x + 12 x^{2}$ . Its place in the bracket is taken by 1.

Worked example

1
$4 x + 12 x^{2}$ — The question. Both terms share a number and an x.
2
$common factor = 4 x$ — 4 is the biggest number that goes into 4 and 12, and the lowest power of x is x.
3
$4 x \div 4 x = 1$ — Divide the first term by 4x. Anything divided by itself is 1, not 0.
4
$12 x^{2} \div 4 x = 3 x$ — Divide the second term: 12 ÷ 4 = 3 and $x^{2} \div x = x$ .
5
$= 4 x (1 + 3 x)$ — Common factor outside, 1 and 3x inside.
6
$4 x \times 1 + 4 x \times 3 x = 4 x + 12 x^{2} ✓$ — Expand to check. Both terms come back, so it is right.

Common mistake

Writing $4 x (3 x)$ . It expands to $12 x^{2}$ only, so the 4x has vanished. Leave 1 in its place.

Find the HCF of Two Terms

Some questions only ask for the highest common factor (HCF) of two terms, like $18 x^{4}$ and $24 x^{7}$ . It is the part you would take outside the bracket, with no bracket.

Worked example

1
$18 x^{4} and 24 x^{7}$ — The question. Deal with the numbers first, then the letter.
2
$18 = 6 \times 3, 24 = 6 \times 4$ — 6 is the biggest number that goes into both 18 and 24.
3
$x^{4}, x^{7} \to x^{4}$ — $x^{4}$ is four x's multiplied, and $x^{7}$ contains those same four. So both share x⁴: the lowest power.
4
$HCF = 6 x^{4}$ — Multiply the pieces together. The answer is one term, with no bracket.

Common mistake

Taking the highest power, $x^{7}$ . $18 x^{4}$ has only four x's, so it cannot share seven.

A Square Minus a Square

Two square terms with a minus between them, like $9 x^{2} - 16 y^{2}$ , share nothing. They factorise into two brackets.

Worked example

1
$9 x^{2} - 16 y^{2}$ — The question. Two terms, a minus between them, and both are squares.
2
$= (3 x)^{2} - (4 y)^{2}$ — Find what is squared: 3x × 3x = 9x², and 4y × 4y = 16y².
3
$= (3 x - 4 y) (3 x + 4 y)$ — Write two brackets, each with 3x and 4y. One bracket gets a minus, the other a plus.

Key idea

Why it works: expand it back and the middle terms, $- 12 x y$ and $+ 12 x y$ , cancel.

“Factorise fully” may need a number out first:

Worked example

1
$3 m^{2} - 75$ — The question. 75 is not a square number, so this is not a square minus a square yet.
2
$= 3 (m^{2} - 25)$ — Take out the common factor 3: $3 m^{2} \div 3 = m^{2}$ and 75 ÷ 3 = 25.
3
$= 3 (m^{2} - 5^{2})$ — Now the bracket is a square minus a square, because 25 = 5 × 5.
4
$= 3 (m - 5) (m + 5)$ — Split the bracket into two brackets, one with − and one with +. Keep the 3 in front.

Common mistake

Stopping at $3 (m^{2} - 25)$ . The bracket still factorises, so it is not fully factorised.

Which Method? Read the Question

Count the terms and look at the signs.

You see	You do
$9 y - 12$	Take out the biggest number
“Fully”, with letters	Biggest number and the lowest power of each letter
One term is the whole common factor	Leave 1 in its place
“Find the HCF”	Biggest number and lowest powers, with no bracket
Square minus square	Two brackets, one with − and one with +
Three terms with x²	Two brackets: see Factorising Quadratic Expressions

Key idea

Expand your answer in your head. If you get the question back, it is right.

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