GCSE Maths Algebra Factorising

Difference of Two Squares

# Difference of Two Squares

Here we will learn how to factorise using the difference of two squares method for a quadratic in the form a2 – b2.

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.

## What is the difference of two squares?

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

a2 – b2

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 x2 – 4 or 4x2 – 25.

To use quadratic factorisation on an expression in the form a2 – b2, we need double brackets. a2 – b2 = (a + b)(a – b) ### What is the difference of two squares? ## How to factorise using difference of two squares

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

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

### Explain how to factorise using the difference of two squares in 4 steps ## Difference of two squares examples

### Example 1: x2 variable has a coefficient of 1

Fully factorise

x2 – 9

1. Write down two brackets

(      )(      )

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

x2 – 9

√x2

(x     )(x     )

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

(x2 – 9)

√9

(x     3)(x     3)

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

(x + 3)(x3)

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

(x + 3)(x3) = x2 – 9

### Example 2: y2 variable has a coefficient of 1

Fully factorise

64 – y2

(        )(        )

64 – y2

√64

(8       )(8       )

64 – y2

√y2

(8     y)(8     y)

(8 + y)(8y)

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

(8y)(8 + y) = 64 – y2

### Example 3: x2 variable has a coefficient greater than 1

Fully factorise

25x2 – 16

(      )(      )

25x2 – 16

√25x2

(5x      )(5x       )

25x2 – 16

√16

(5x    4)(5x    4)

(5x + 4)(5x4)

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

(5x + 4)(5x4) = 25x2 – 16

### Example 4: coefficient of both terms is greater than 1

Fully factorise

4x2 – 81y2

(      )(      )

4x2 – 81y2

√4x2

(2x      )(2x       )

4x2 – 81y2

√81y2

(2x   9y)(2x    9y)

(2x + 9y)(2x9y)

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

(2x + 9y)(2x9y) = 4x2 – 81y2

### Example 5: combining HCF / GCF and difference of two squares

Fully factorise

x3 – 64x

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

We first need to find the highest or greatest common factor (x) and write it outside of a single bracket.

x(x2 – 64)

x(      )(      )

x(x2 – 64)

√x2

x(x      )(x       )

x(x2 – 64)

√64

x(x   8)(x     8)

x(x + 8)(x8)

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

x(x + 8)(x8) = x3 – 64x

## Common misconceptions

• The order of subtraction

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

(    +   )(    –    ) or (   –   )(   +   )

• Square root the entire term

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

E.g.

9x2 – 49

√9x2 = 3x

√49 = 7

• Terms used can vary

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

## Practice difference of two squares questions

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

(4x+10)(4x-10) (4x-10)(4x-10) (16x+100)(x-1) (16x+1)(x-100) Recognising the expression as a difference of two squares means we can square root each term

\begin{aligned} \sqrt{16x^{2}}&=4x\\ \sqrt{100}&=10\\ \end{aligned}

and then rewrite as a product of two brackets

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

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)

## Difference of two squares GCSE questions

1. Factorise: x2 – 100

= (x + 10)(x – 10)

(2 marks)

2. Factorise: y2 – 49

= (y + 7)(y – 7)

(2 marks)

3. Factorise: 2x2 – 50

= 2(x2 – 25) = 2(x + 5)(x – 5)

(2 marks)

### Learning checklist

You have now learnt to:

• Manipulate algebraic expressions by taking out common factors to factorise into a single bracket
• Factorise quadratic expressions of the form x2 + bx + c
• Factorise quadratic expressions of the form of the difference of two squares
• Factorising quadratic expressions of the form ax2 + bx + c (H)

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