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Here we will learn about factorising quadratics; we will explore what quadratic expressions are and the steps needed to factorise into double brackets.

There are also factorising quadratics worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if youβre still stuck.

First of all, letβs have a quick recap on quadratic expressions.

A quadratic expression in maths is an expression including a squared term or square number i.e. a term up to ^{2}

The highest power for a quadratic expression is

The general form of a quadratic expression is:

\[\color{#FE47EC}ax^{2}\color{#00BC89}{+b}x\color{#7C4DFF}{+c}\]

^{2}

e.g.

\[x^{2}\color{#00BC89}{-2}x\color{#7C4DFF}{+1}\]

\[\color{#FE47EC}2x^{2}\color{#00BC89}{+3}x\color{#7C4DFF}{-2}\]

We factorise quadratic expressions of this sort using double brackets. There are different methods we can use depending on whether the coefficient of ^{2}

Factorising, or factoring quadratic equations is the opposite of expanding brackets and is used to solve quadratic equations.

For example, in the form of ^{2} + bx + c

**How to factorise quadratics: **

- Write out the factor pairs of the last number (c).
- Find a pair of factors that
**+**to give the middle number (b) and multiply to give the last number (c). - Write two brackets and put the variable at the start of each one.
- Write one factor in the first bracket and the other factor in the second bracket. The order isnβt important, but the signs of the factors are.

If you’re looking for a summary of all the different ways you can factorise expressions then you may find it helpful to start with our main factorising lesson or look in detail at the other lessons in this section.

A quadratic equation is a quadratic expression that is equal to something. We can solve quadratic equations by using factorisation (or factoring), the quadratic formula or by completing the square.

**Step by step guide: **Quadratic equations

Download two free factorising quadratics worksheets to help your students prepare for GCSEs. Includes reasoning and applied questions.

DOWNLOAD FREEDownload two free factorising quadratics worksheets to help your students prepare for GCSEs. Includes reasoning and applied questions.

DOWNLOAD FREETo factorise a quadratic expression in the form ^{2} + bx + c**double brackets**. Factorisation into double brackets is the reverse process of expanding double brackets.

In this case, the coefficient (number in front) of the ^{2}

In order to factorise a** quadratic** algebraic expression in the form ^{2} + bx + c

- Write out the factor pairs of the last number
(c) . - Find a pair of factors that
**+**to give the middle number(b) and β to give the last number(c) . - Write two brackets and put the variable at the start of each one.
- Write one factor in the first bracket and the other factor in the second bracket. The order isnβt important, the signs of the factors are.

Fully factorise:

\[x^2 \color{#00BC89}{+ 6x}\color{#7C4DFF} {+ 5}\]

- Write out the factor pairs of the last number (5) in order.

\[x^2 \color{#00BC89}{+ 6x}\color{#7C4DFF} {+ 5}\]

Factors of 5:

1, 5

2We need a pair of factors that + to give the middle number (6) and β to give the last number (5).

\[x^2 \color{#00BC89}{+ 6x}\color{#7C4DFF} {+ 5}\]

Factors of 5:

1, 5

1 + 5= 6β

1 β 5= 5β

(Itβs a good idea to do a quick check that we have the correct numbers)

Remember: to multiply two values together to give a positive answer, the signs must be the same

3Write two brackets and put the variable at the start of each one (x in this case).

\[(x\qquad)(x\qquad)\]

4Write one factor in the first bracket and the other factor in the second bracket. The order isnβt important, the signs of the factors are.

\[(x+\color{#FF9100}1)(x+\color{#FF9100}5)\]

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

\[(x+1)(x+5)=x^{2}+6x+5\]

Fully factorise:

\[x^2 – 2x – 24\]

Write out the factor pairs of the last number (24) in order

^{2} – 2x – 24

Factors of 24 :

\begin{aligned} &1, 24 \\ &2, 12 \\ &3, 8 \\ &4, 6 \end{aligned}We need a pair of factors that + to give the middle number (-2) and β to give the last number (-24).

^{2} – 2x – 24

Factors of 24:

1, 24

2, 12

3, 8

4, 6

-6 + 4 = -2 β

-6 β 4 = -24 β

(Itβs a good idea to do a quick check that we have the correct numbers)

Remember: to multiply two values together to give a negative answer, the signs must be the different.

Write two brackets and put the variable at the start of each one (x in this case).

Write one factor in the first bracket and the other factor in the second bracket. The order isnβt important, the signs of the factors are.

(x – 6)(x + 4)

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

^{2} – 2x – 24

Fully factorise:

^{2} + x – 20

Write out the factor pairs of last number (20) in order.

\[x^{2}+x-20\]

Factors of 20:

1, 20

2, 10

4, 5

We need a pair of factors that + to give the middle number (1) and β to give the last number (-20).

^{2} + x – 20

Factors of 20:

1, 20

2, 10

4, 5

-4 + 5 = 1 β

-4 β 5 = -20 β

(Itβs a good idea to do a quick check that we have the correct numbers)

Remember: to β two values together to give a negative answer, the signs must be the different

Write two brackets and put the variable at the start of each one (x in this case)

(x – 4)(x + 5)

We can check the answer by multiplying out the brackets!

^{2} + x – 20

Fully factorise:

\[x^2 – 8x + 15\]

Write out the factor pairs of the last number (15) in order.

Factors of 15:

1, 15

3, 5

We need a pair of factors that + to give the middle number (-8) and β to give the last number (15).

^{2} – 8x + 15

Factors of 15:

1, 15

3, 5

-3 + -5 = -8 β

-3 β -5 = 15 β

Itβs a good idea to do a quick check that we have the correct numbers.

Remember: to β two values together to give a positive answer, the signs must be the same

Write two brackets and put the variable at the start of each one (x in this case)

Write one factor in the first bracket and the other factor in the second bracket.

\[(x – 3)(x – 5)\]

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

^{2} – 8x + 15

1. Fully factorise:

x^{2}+5x+6

(x+5)(x+6)

(x+3)(x+2)

(x+1)(x+6)

(x+3)(x+3)

We need numbers that have a product of 6 and a sum of 5 . By considering the factors of 6 , we conclude that x^{2}+5x+6=(x+3)(x+2) .

2. Fully factorise:

x^{2}+10x+21

(x+3)(x+7)

(x+1)(x+21)

(x+10)(x+21)

(x+1)(x+10)

We need numbers that have a product of 21 and a sum of 10 . By considering the factors of 21 , we conclude that x^{2}+10x+21=(x+3)(x+7) .

3. Fully factorise:

x^{2}-x-12

(x-4)(x+3)

(x+4)(x-3)

(x-1)(x+12)

(x-1)(x-12)

We need numbers that have a product of -12 and a sum of -1 . By considering the factors of -12 , we conclude that x^{2}-x-12=(x-4)(x+3) .

4. Fully factorise:

x^{2}+3x-18

(x+2)(x-9)

(x+6)(x-3)

(x+3)(x-6)

(x-6)(x-3)

We need numbers that have a product of -18 and a sum of 3 . By considering the factors of -18 , we conclude that x^{2}+3x-18=(x+6)(x-3) .

5. Fully factorise:

x^{2}-6x+8

(x-2)(x-6)

(x-2)(x+4)

(x+2)(x-4)

(x-2)(x-4)

We need numbers that have a product of 8 and a sum of -6 . By considering the negative factors of 8 , we conclude that x^{2}-6x+8=(x-2)(x-4) .

6. Fully factorise:

x^{2}-10x+24

(x-4)(x-6)

(x-2)(x-12)

(x+4)(x-6)

(x-4)(x+6)

We need numbers that have a product of 24 and a sum of -10 . By considering the negative factors of 24 , we conclude that x^{2}-10x+24=(x-4)(x-6) .

1. Factorise: x^{2} + 3x – 10

Show answer

(x – 2)(x + 5)

(2 marks)

2. Factorise: y^{2} – 10y + 16

Show answer

(y – 2)(y – 8)

(2 marks)

3. Factorise: x^{2} – 12x + 27

Show answer

(x – 3)(x – 9)

(2 marks)

To factorise a quadratic expression in the form ^{2} + bx + c**double brackets**. Factorising into double brackets is the reverse process of expanding double brackets.

In this case the coefficient (number in front) of the ^{2}

In order to factorise a quadratic algebraic expression in the form ^{2} + bx + c

- Multiply the end numbers together (
a andc ) then write out the factor pairs of this new number in order. - We need a pair of factors that + to give the middle number (
b ) and β to give this new number. - Rewrite the original expression, this time splitting the middle term into the two factors we found in step 2. The order of these factors doesnβt matter, the signs do.
- Split the equation down the middle and fully factorise each half. The expressions in the brackets must be the same!
- Factorise the whole expression by bringing the contents of the bracket to the front and writing the two other terms in the other bracket.

Fully factorise:

\[2x^2 + 5x + 3\]

- Multiply the end numbers together (2 and 3) then write out the factor pairs of this new number in order.

^{2} + 5x + 3

Factors of 6:

1, 6

2, 3

2We need a pair of factors that + to give the middle number (

^{2} + 5x + 3

Factors of 6:

1, 6

2, 3

+ 5

β 6

2 + 3 = 5 β

2 x 3 = 6 β

Remember: to x two values together to give a positive answer, the signs must be the same.

3Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesnβt matter, the signs do.

2x ^{2}+ 5x + 3

2x ^{2}+ 2x + 3x + 3

4Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!

2x ^{2}+ 5x + 3

2x ^{2}+ 2x + 3x + 3

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

**(x + 1)** + 3**(x + 1)**

5Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket.

(x + 1) ( 2x \; + \; 3 )

The order of the brackets doesnβt matter

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

^{2} + 5x + 3

Fully factorise:

\[2x^2 + 3x – 2\]

Multiply the the end numbers together (2 and -2) then write out the factor pairs of this new number in order.

^{2} + 3x – 2

Factors of 4:

1, 4

2, 2

We need a pair of factors that + to give the middle number (3) and β to give this new number (-4)

^{2} + 3x – 2

Factors of 4:

1, 4

2, 2

β 3

β -4

-1 + 4 = 3 β

-1 β 4 = -4 β

Remember: to x two values together to give a negative answer, the signs must be different

Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesnβt matter, the signs do.

2x ^{2}+ 3x - 2

2x ^{2}- x + 4x - 2

Split the equation down the middle into two halves and fully factorise each half β the expressions in the brackets must be the same!

2x^{2}+ 3x - 2 2x^{2}- x + 4x - 2 x(2x + 1) + 2(2x - 1)

Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket.

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

^{2} + 3x – 2

Fully factorise:

\[3x^2 – 2x – 8\]

Multiply the the end numbers together (3 and -8) then write out the factor pairs of this new number in order.

^{2} – 2x – 8

Factors of 24:

1, 24

2, 12

3, 8

4, 6

We need a pair of factors that + to give the middle number (-2) and to β give this new number (-24)

^{2} – 2x – 8

Factors of 24:

1, 24

2, 12

3, 8

4, 6

β -2

β -24

-6 + 4 = -2 β

-6 β 4 = -24 β

Remember: to β two values together to give a negative answer, the signs must be different

3x ^{2}- 2x - 8

3x ^{2}- 6x + 4x - 8

Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!

3x ^{2}- 2x - 8

3x ^{2}- 6x + 4x - 8

3x (x - 2)+ 4(x - 2)

(x – 2)(3x + 4)

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

^{2} – 2x – 8

Fully factorise:

\[ 6x^2 – 7x + 2 \]

Multiply the the end numbers together (6 and 2) then write out the factor pairs of this new number in order.

^{2} – 7x + 2

Factors of 12:

1, 12

2, 6

3, 4

We need a pair of factors that + to give the middle number (-7) and β to give this new number (12)

^{2} – 7x + 2

Factors of 12:

1, 12

2, 6

3, 4

+ -7

β -24

-3 + -4 = -7 β

-3 β -4 = 12 β

Remember: to β two values together to give a positive answer, the signs must be the same

6x ^{2}- 7x + 2

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

6x ^{2}- 7x + 2

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

3x (2x - 1)- 2(2x - 1)

(2x – 1)(3x – 2)

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

^{2} – 7x + 2

1. Fully factorise:

2x^{2}+5x+2

(x+1)(2x+2)

(2x+1)(x+2)

(2x+2)(x+2)

(2x+5)(x+2)

Using the method from the above lesson, we can rewrite 2x^{2}+5x+2 as 2x^{2}+4x+x+2 which can be factorised as 2x(x+2)+1(x+2) or more concisely, (2x+1)(x+2) .

2. Fully factorise:

2x^{2}+x-6

(2x-3)(x+2)

(2x+3)(x+2)

(2x-3)(x-2)

(2x-1)(x+6)

Using the method from the above lesson, we can rewrite 2x^{2}+x-6 as 2x^{2}-3x+4x-6 which can be factorised as x(2x-3)+2(2x-3) or more concisely, (2x-3)(x+2) .

3. Fully factorise:

2x^{2}-14x+20

(2x-4)(x-5)

2(x-2)(x-5)

(2x-1)(x-20)

(x-4)(2x-5)

For the given expression, 2x^{2}-14x+20 first factor out the HCF of 2 , giving 2[x^{2}-7x+10] then use the method from the lesson to factorise this simpler expression, hence 2[x^{2}-5x-2x+10] or 2[x(x-5)-2(x-5)] so that the fully factorised expression is 2(x-2)(x-5) .

4. Fully factorise:

3x^{2}-7x-6

(3x+3)(x-2)

(3x-2)(x+3)

(3x+2)(x-3)

(x+2)(3x-3)Β

Using the method from the above lesson, we can rewrite 3x^{2}-7x-6 as 3x^{2}-9x+2x-6 which can be factorised as 3x(x-3)+2(x-3) or more concisely, (3x+2)(x-3) .

5. Fully factorise:

3x^{2}-7x+2

(3x+1)(x-2)

(3x-1)(x+2)

(x-1)(3x-2)

(3x-1)(x-2)

Using the method from the above lesson, we can rewrite 3x^{2}-7x+2 as 3x^{2}-6x-x+2 which can be factorised as 3x(x-2)-(x-2) or more concisely, (3x-1)(x-2) .

6. Fully factorise:

4x^{2}-18x+8

(4x+2)(x-4)

(4x-2)(x+4)

2(2x-1)(x-4)

(4x-2)(x-4)

For the given expression, 4x^{2}-18x+8 first factor out the HCF of 2 , givingΒ 2(2x^{2}-9x+4) then use the method from the lesson to factorise this simpler expression, hence 2[2x^{2}-8x-x+4] or 2[2x(x-4)-(x-4)] so that the fully factorised expression is 2(2x-1)(x-4) .

1. Factorise: 2x^{2} + 9x + 4

Show answer

(2x + 1)(x + 4)

(2 marks)

2. Factorise: 2y^{2} – y – 3

Show answer

(2y – 3)(y + 1)

(2 marks)

3. Factorise: 2x^{2} – x – 10

Show answer

(2x – 5)(x + 2)

(2 marks)

**The order of the brackets**

When we multiply two values the order doesnβt matter. This is true for the brackets when factorising quadratics

e.g.2 Γ 3 = 3 Γ 2

It is exactly the same here.(x – 6)(x + 4) means(x – 6)(x + 4)

So,(x – 6)(x + 4)=(x + 4)(x – 6)

**The signs of the factors**

It is common for students to get confused regarding the signs of the factors in the brackets, especially with negative factors.

e.g. if the factors are-6 and4 , the numbers is the brackets must be:(x – 6)(x + 4)

**Multiplying two numbers to give a +**

For two numbers to multiply to give a + their signs must be the same, that is a negative x a negative or a positive x a positive.

+ Γ + = +

e.g 2 Γ 3 = 6

4 Γ 5=20

– Γ – = +

e.g -2 Γ -3 = 6

-4 Γ -5= 20

**Multiplying two numbers to give a**–

For two numbers to multiply to give a – their signs must be different, that is a negative x a positive or a positive x a negative.

+ Γ – = –

e.g 2 Γ -3= -6

4 Γ -5= -20

– Γ + = –

e.g -2 Γ 3= -6

-4 Γ 5= -20

**The signs are important for the end numbers as well**

e.g. 3x^{2} - 2x - 8

3 x -8 = -24 **NOT 24**

**See also:**Negative numbers

You have now learnt how to:

- Manipulate algebraic expressions by taking out common factors to factorise into a single bracket
- Factorise quadratic expressions of the form x^{2} + bx + c
- Factorise quadratic expressions of the form of the difference of two squares
- Factorising quadratic expressions of the form ax^{2} + bx + c (H)

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