Factorising to a single bracket is the
opposite of expanding a single
bracket. You should be familiar with lessons 5 and 8 before starting,
and be able to work out the highest common factor (HCF) of two or three numbers.
Oh, and be prepared for a bit of a challenge.

SECTION A

Expand \(~4(2x+3)~\) and you get \(~8x+12~\). Factorise \(~8x+12~\) and you get \(~4(2x+3)~\).
That's what we mean when we say expanding and factorising are opposites. But
how do we get from \(~8x+12~\) to \(~4(2x+3)~\)?

Example 1

Factorise fully \(~8x+12~\)

First find the highest common factor (HCF) of the numbers 8 and 12

The HCF is 4 so the answer will look something like this ⇒ \(~4(■~~~~□)~\)

If we expand the answer, we want to get \(~8x+12~\) so ...

\(~4\times ■=8x~\) ⇒ so \(~■~\) must be \(~2x~\)

\(~4\times □=12~\) ⇒ so \(~□~\) must be \(~3~\)

Answer: \(~4(2x+3)~\)

Important!

The example above says factorise fully \(~8x+12~\).
What does this mean? Well, let's say you just found any old factor of the two numbers instead of the
highest common factor. Let's say you came up with 2.

\(~2\times ■=8x~\) ⇒ so \(~■~\) must be \(~4x~\)

\(~2\times □=12~\) ⇒ so \(~□~\) must be \(~6~\)

Answer: \(~2(4x+6)~\)

You have factorised \(~8x+12~\) but you have not fully factorised it.
You must always fully factorise even if the question just says factorise!

Checking your answer

If you expand your fully factorised answer \(~4(2x+3)~\), it should give you \(~8x+12~\).
This is a useful check to make sure you have factorised correctly.

BUT if you expand \(~2(4x+6)~\), you will also get \(~8x+12~\)!
So this check will pick up on most mistakes, which you can then correct,
but it won't tell you if you've fully
factorised or not. So be careful, find the HCF, not just any old factor!

One way of checking that you've fully
factorised is to look inside the brackets in your answer. Could this be factorised further?
If so, you ain't finished!

Example 2

Factorise \(~12-18x~\)

First find the highest common factor (HCF) of the numbers 12 and 18

The HCF is 6 so the answer will look something like this ⇒ \(~6(■~~~~□)~\)

If we expand the answer, we want to get \(~12-18x~\) so ...

\(~6\times ■=12~\) ⇒ so \(~■~\) must be \(~2~\)

\(~6\times □=-18x~\) ⇒ so \(~□~\) must be \(~-3x~\)

Answer: \(~6(2-3x)~\)

Now it's your turn!

Practise to master

SECTION A

Factorise the following expressions.

01) \(~6x+14~\)

HCF of 6 and 14 ⇒ 2 \(2(3x+7)\)

02) \(~7x+21~\)

HCF of 7 and 21 ⇒ 7 \(7(x+3)\)

03) \(~10x-25~\)

HCF of 10 and 25 ⇒ 5 \(5(2x-5)\)

04) \(~24x-6~\)

HCF of 24 and 6 ⇒ 6 \(6(4x-1)\)

05) \(~8x+36~\)

HCF of 8 and 36 ⇒ 4 \(4(2x+9)\)

06) \(~5x+20~\)

HCF of 5 and 20 ⇒ 5 \(5(x+4)\)

07) \(~15-9x~\)

HCF of 15 and 9 ⇒ 3 \(3(5-3x)\)

08) \(~9-27x~\)

HCF of 9 and 27 ⇒ 9 \(9(1-3x)\)

SECTION B

In Section A, you looked at the number part of each term and worked
out the HCF. There was also an \(~x~\) part but only in one of the terms.

In this section, both terms will have a number part and
both terms will have an \(~x~\) part so we'll have to work out a HCF for each.

Example 1

Factorise fully \(~6x^2+8x~\)

First find the HCF of 6 and 8 ⇒ HCF is 2

Now find the HCF of \(~x^2~\) and \(~x~\) ⇒ HCF is \(~x~\) (see Tip! below)

So the answer will look something like this ⇒ \(~2x(■~~~~□)~\)

If we expand the answer, we want to get \(~6x^2+8x~\) so ...

\(~2x\times ■=6x^2~\) ⇒ so \(~■~\) must be \(~3x~\)

\(~2x\times □=8x~\) ⇒ so \(~□~\) must be \(~4~\)

Answer: \(~2x(3x+4)~\)

Tip!
The HCF of the \(~x~\) bits is just the one with the lowest power. Simple as that!
For example, the HCF of \(~x^5~\) and \(~x^2~\) is \(~x^2~\), the HCF of \(~y^3~\) and \(~y~\) is \(~y~\), etc.

It usually takes a good few practice questions before students start to get a feel for this.

Practise to master

SECTION B

Factorise the following expressions.

01) \(~3x^2+7x~\)

HCF of 3 and 7 ⇒ 1 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(x(3x+7)\)

02) \(~4x^2+x~\)

HCF of 4 and 1 ⇒ 1 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(x(4x+1)\)

03) \(~4x^2+6x~\)

HCF of 4 and 6 ⇒ 2 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(2x(2x+3)\)

04) \(~6x^2+18x~\)

HCF of 6 and 18 ⇒ 6 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(6x(x+3)\)

05) \(~5x^2-6x~\)

HCF of 5 and 6 ⇒ 1 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(x(5x-6)\)

06) \(~x^2+2x~\)

HCF of 1 and 2 ⇒ 1 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(x(x+2)\)

07) \(~8x^2-12x~\)

HCF of 8 and 12 ⇒ 4 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(4x(2x-3)\)

08) \(~35x^2+7x~\)

HCF of 35 and 7 ⇒ 7 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(7x(5x+1)\)

09) \(~4x^2+9x^3~\)

HCF of 4 and 9 ⇒ 1 HCF of \(~x^2~\) and \(~x^3~\) ⇒ \(~x^2~\) \(x^2(4+9x)\)

10) \(~2x^3-x~\)

HCF of 2 and 1 ⇒ 1 HCF of \(~x^3~\) and \(~x~\) ⇒ \(~x~\) \(x(2x^2-1)\)

11) \(~9x^2+24x~\)

HCF of 9 and 24 ⇒ 3 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(3x(3x+8)\)

12) \(~5x^4+30x^2~\)

HCF of 5 and 30 ⇒ 5 HCF of \(~x^4~\) and \(~x^2~\) ⇒ \(~x^2~\) \(5x^2(x^2+6)\)

13) \(~7x-10x^5~\)

HCF of 7 and 10 ⇒ 1 HCF of \(~x~\) and \(~x^5~\) ⇒ \(~x~\) \(x(7-10x^4)\)

14) \(~x-5x^2~\)

HCF of 1 and 5 ⇒ 1 HCF of \(~x~\) and \(~x^2~\) ⇒ \(~x~\) \(x(1-5x)\)

15) \(~15x^2-20x~\)

HCF of 15 and 20 ⇒ 5 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) \(5x(3x-4)\)

16) \(~21x^2-3x^3~\)

HCF of 21 and 3 ⇒ 3 HCF of \(~x^2~\) and \(~x^3~\) ⇒ \(~x^2~\) \(3x^2(7-x)\)

SECTION C

You've done well to get this far! They say stretching is good for you. This
last section will certainly do that!

Example 1

Factorise \(~8x^2y^2+20xy^4~\)

HCF of 8 and 20 ⇒ 4

HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\)

HCF of \(~y^2~\) and \(~y^4~\) ⇒ \(~y^2~\)

So the answer will look something like this ⇒ \(~4xy^2(■~~~~□)~\)

When we expand the answer, we want to get \(~8x^2y^2+20xy^4~\) so ...

\(~4xy^2\times ■=8x^2y^2~\) ⇒ so \(~■~\) must be \(~2x~\)

\(~4xy^2\times □=20xy^4~\) ⇒ so \(~□~\) must be \(~5y^2~\)

Answer: \(~4xy^2(2x+5y^2)~\)

Example 2

Factorise \(~2xy^3+8x^5y-4y^2~\)

HCF of 2, 8 and 4 ⇒ 2

There isn't an \(~x~\) part in every term!

HCF of \(~y^3~\), \(~y~\) and \(~y^2~\) ⇒ \(~y~\)

So the answer will look something like this ⇒ \(~2y(■~~~~□~~~~▲)~\)

When we expand the answer, we want to get \(~2xy^3+8x^5y-4y^2~\) so ...

\(~2y\times ■=2xy^3~\) ⇒ so \(~■~\) must be \(~xy^2~\)

\(~2y\times □=8x^5y~\) ⇒ so \(~□~\) must be \(~4x^5~\)

\(~2y\times ▲=-4y^2~\) ⇒ so \(~▲~\) must be \(~-2y~\)

Answer: \(~2y(xy^2+4x^5-2y)~\)

The following questions will require quite a bit of concentration.

Practise to master

SECTION C

Factorise the following expressions.

01) \(~2y+5xy^2~\)

HCF of 2 and 5 ⇒ 1 There isn't an \(~x~\) part in both terms! HCF of \(~y~\) and \(~y^2~\) ⇒ \(~y~\) \(y(2+5xy)\)

02) \(~7xy^3-x~\)

HCF of 7 and 1 ⇒ 1 HCF of \(~x~\) and \(~x~\) ⇒ \(~x~\) There isn't a \(~y~\) part in both terms! \(x(7y^3-1)\)

03) \(~8ab^2+10a^2b~\)

HCF of 8 and 10 ⇒ 2 HCF of \(~a~\) and \(~a^2~\) ⇒ \(~a~\) HCF of \(~b^2~\) and \(~b~\) ⇒ \(~b~\) \(2ab(4b+5a)\)

04) \(~3x^2y^2-9xy^3~\)

HCF of 3 and 9 ⇒ 3 HCF of \(~x^2~\) and \(~x~\) ⇒ \(~x~\) HCF of \(~y^2~\) and \(~y^3~\) ⇒ \(~y^2~\) \(3xy^2(x-3y)\)

05) \(~7y+5x^2y^3~\)

HCF of 7 and 5 ⇒ 1 There isn't an \(~x~\) part in both terms! HCF of \(~y~\) and \(~y^3~\) ⇒ \(~y~\) \(y(7+5x^2y^2)\)

06) \(~p^2q^2-6q~\)

HCF of 1 and 6 ⇒ 1 There isn't a \(~p~\) part in both terms! HCF of \(~q^2~\) and \(~q~\) ⇒ \(~q~\) \(q(p^2q-6)\)

07) \(~4xy+14x^3~\)

HCF of 4 and 14 ⇒ 2 HCF of \(~x~\) and \(~x^3~\) ⇒ \(~x~\) There isn't a \(~y~\) part in both terms! \(2x(2y+7x^2)\)

08) \(~10x^3y^2-5xy^2~\)

HCF of 10 and 5 ⇒ 5 HCF of \(~x^3~\) and \(~x~\) ⇒ \(~x~\) HCF of \(~y^2~\) and \(~y^2~\) ⇒ \(~y^2~\) \(5xy^2(2x^2-1)\)

09) \(~6xy^2+4xy+8xy^3~\)

HCF of 6, 4 and 8 ⇒ 2 HCF of \(~x~\), \(~x~\) and \(~x~\) ⇒ \(~x~\) HCF of \(~y^2~\), \(~y~\) and \(~y^3~\) ⇒ \(~y~\) \(2xy(3y+2+4y^2)\)

10) \(~3y^2+9x^2y-6x^3y~\)

HCF of 3, 9 and 6 ⇒ 3 There isn't an \(~x~\) part in every term! HCF of \(~y^2~\), \(~y~\) and \(~y~\) ⇒ \(~y~\) \(3y(y+3x^2-2x^3)\)

11) \(~x^2y^2-4y+12xy~\)

HCF of 1, 4 and 12 ⇒ 1 There isn't an \(~x~\) part in every term! HCF of \(~y^2~\), \(~y~\) and \(~y~\) ⇒ \(~y~\) \(y(x^2y-4+12x)\)

12) \(~8xy-2x^2-4x^2y^2~\)

HCF of 8, 2 and 4 ⇒ 2 HCF of \(~x~\), \(~x^2~\) and \(~x^2~\) ⇒ \(~x~\) There isn't a \(~y~\) part in every term! \(2x(4y-x-2xy^2)\)