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In order to access this I need to be confident with:

Expanding brackets Expand and simplifyFactors, multiples, powers and roots

Algebraic expressionsAdding and subtracting negative numbers

Multiplying and dividing negative numbers

Highest common factor (HCF)

Law of indices Square numbers and square rootsThis topic is relevant for:

Here we will learn how to factorise using the difference of two squares method for a quadratic in the form ^{2} – b^{2}

There are also difference of two squares questions and an extensive worksheet based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

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.

The difference of two squares is a method of factorising used when an algebraic expression includes two squared terms, one subtracted from the other:

^{2} – b^{2}

When we are subtracting a squared term from another squared term we can use the difference of two squares method. Square numbers are sometimes called perfect squares.

An example of an expression we can factorise using the difference of two squares might be ^{2} – 4 or 4x^{2} – 25

To use quadratic factorisation on an expression in the form ^{2} – b^{2}

Get your free difference of two squares worksheet with 20+ reasoning and applied questions, answers and mark scheme.

DOWNLOAD FREEGet your free difference of two squares worksheet with 20+ reasoning and applied questions, answers and mark scheme.

DOWNLOAD FREEIn order to factorise an algebraic expression using the difference of two squares:

- Write down two brackets.
- Square root the first term and write it on the left hand side of both brackets.
- Square root the last term and write it on the right hand side of both brackets.
- Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter.)

Fully factorise

^{2} – 9

- Write down two brackets

2Square root the first term and write it on the left hand side of both brackets

^{2} – 9

^{2})

3Square root the last term and write it on the right hand side of both brackets.

^{2} – 9)

4Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter).

We can check the final answer by multiplying out the brackets!

^{2} – 9

Fully factorise

^{2}

Write down two brackets

Square root the first term and write it on the left hand side of both brackets

^{2}

Square root the last term and write it on the right hand side of both brackets

^{2}

^{2})

Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter.)

We can check the final answer by multiplying out the brackets!

^{2}

Fully factorise

^{2} – 16

Write down two brackets

Square root the first term and write it on the left hand side of both brackets

^{2} – 16

^{2}

Square root the last term and write it on the right hand side of both brackets

^{2} – 16

Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter.

We can check the final answer by multiplying out the brackets!

^{2} – 16

Fully factorise

^{2} – 81y^{2}

Write down two brackets

Square root the first term and write it on the left hand side of both brackets

^{2} – 81y^{2}

^{2})

Square root the last term and write it on the right hand side of both brackets

^{2} – 81y^{2}

^{2})

Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter.)

We can check the final answer by multiplying out the brackets!

^{2} – 81y^{2}

Fully factorise

^{3} – 64x

*Be careful, this one is not the difference of two squares!*

We first need to find the highest or greatest common factor (

^{2} – 64)

Write down two brackets with the x at the front

Square root the first term and write it on the left hand side of both brackets

^{2} – 64)

^{2})

Square root the last term and write it on the right hand side of both brackets

^{2} – 64)

Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter.)

We can check the final answer by multiplying out the brackets!

^{3} – 64x

**The order of subtraction**

There must simply be a + in one bracket, and a – in the other, the order doesn’t matter.

**Square root the entire term**

Remember to square root the entire term including the coefficients of the variables.

E.g.^{2} – 49^{2}) = 3x

**Terms used can vary**

The term factorising can sometimes be written as ‘factoring’ or factorization’

1. Fully factorise:

x^{2}-25

(x-5)(x-5)

(x+5)(x-5)

(x+1)(x-25)

(x-1)(x+25)

Recognising the expression as a difference of two squares means we can square root each term

\begin{aligned} \sqrt{x^{2}}&=x\\ \sqrt{25}&=5\\ \end{aligned}

and then rewrite as a product of two brackets

(x+5)(x-5)

2. Fully factorise:

y^{2}-81

(y+9)(y-9)

(y-9)(y-9)

(y+1)(y-81)

(y-1)(y+81)

Recognising the expression as a difference of two squares means we can square root each term

\begin{aligned} \sqrt{y^{2}}&=y\\ \sqrt{81}&=9\\ \end{aligned}

and then rewrite as a product of two brackets

(y+9)(y-9)

3. Fully factorise:

49-y^{2}

(7-y)(7-y)

(49+y)(1-y)

(7+y)(7-y)

(1+y)(49-y)

Recognising the expression as a difference of two squares means we can square root each term

\begin{aligned} \sqrt{49}&=7\\ \sqrt{y^{2}}&=y\\ \end{aligned}

and then rewrite as a product of two brackets

(7+y)(7-y)

4. Fully factorise:

4-x^{2}

(4+x)(1-x)

(x+2)(x-2)

(1+x)(4-x)

(2+x)(2-x)

Recognising the expression as a difference of two squares means we can square root each term

\begin{aligned} \sqrt{4}&=2\\ \sqrt{x^{2}}&=x\\ \end{aligned}

and then rewrite as a product of two brackets

(2+x)(2-x)

5. Factorise:

16x^{2}-100

4(2x+5)(2x-5)

(4x-10)(4x+10)

(16x+100)(x-1)

(16x+1)(x-100)

First, we can factor 4 out of each term.

This gives 4(4x^{2}-25)

Recognising the expression in the brackets as a difference of two squares means we can square root each term

\begin{aligned} \sqrt{4x^{2}}&=2x\\ \sqrt{25}&=5\\ \end{aligned}

and then rewrite as a product of two brackets muliplied by the 4.

4(2x+5)(2x-5)6. Fully factorise:

49-9y^{2}

(7-3y)(7-3y)

(49+y)(1-9y)

(7+9y)(7-9y)

(7+3y)(7-3y)

Recognising the expression as a difference of two squares means we can square root each term

\begin{aligned} \sqrt{49}&=7\\ \sqrt{9y^{2}}&=3y\\ \end{aligned}

and then rewrite as a product of two brackets

(7+3y)(7-3y)

7. Factorise:

81x^{2}-16y^{2}

(81x+y)(x-4y)

(x+4y)(81x-y)

(9x+4y)(9x-4y)

(9x-4y)(9x-4y)

Recognising the expression as a difference of two squares means we can square root each term

\begin{aligned} \sqrt{81x^{2}}&=9x\\ \sqrt{16y^{2}}&=4y\\ \end{aligned}

and then rewrite as a product of two brackets

(9x+4y)(9x-4y)

8. Fully factorise:

4x^{3}-36x

4x(x+6)(x-6)

2x(x+9)(x-9)

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

4x(x-3)(x-3)

The first thing we need to do is divide by the highest common factor of the two terms and rewrite the expression

4x^{3}-36x=4x(x^{2}-9)

Recognising the expression in the bracket as a difference of two squares means we can square root each term

\begin{aligned} \sqrt{x^{2}}&=x\\ \sqrt{9}&=3\\ \end{aligned}

and then rewrite as a product of two brackets. Don’t forget to also multiply by the factor we initially divided by

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

1. Factorise: x^{2} – 100

Show answer

= (x + 10)(x – 10)

(1 mark)

2. Factorise: y^{2} – 49

Show answer

= (y + 7)(y – 7)

(1 mark)

3. Factorise: 2x^{2} – 50

Show answer

= 2(x^{2} – 25) = 2(x + 5)(x – 5)

(2 marks)

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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