Lesson 5 ~35 min Unit 2 · Algebra +85 XP

Difference of Two Squares

Every expression of the form $a^2 - b^2$ hides a pair of brackets: $(a+b)(a-b)$. Learn to spot perfect squares, take square roots, and factorise in one or two steps.

Today's hook: A builder lays a square patio of side $x$ metres, then cuts out a smaller square fountain of side $3$ metres. The leftover area is $x^2 - 9$ — and it factors neatly into two rectangles.

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Think First

warm-up

Before you read on — factorise $x^2 - 16$. What two numbers multiply to give $-16$ and add to give $0$? Or is there a faster pattern? Try it, then check your reasoning as you go.

Record your answer in your workbook.

The Big Idea

+5 XP to read

When you expand $(a + b)(a - b)$, the middle terms cancel. What remains is $a^2 - b^2$ — a difference of two perfect squares. This pattern is a shortcut: if you see $a^2 - b^2$, you can write it as $(a+b)(a-b)$ instantly.

An expansion of $(a+b)(a-b)$ produces $a^2 - ab + ab - b^2$. The $-ab$ and $+ab$ cancel, leaving $a^2 - b^2$. Factorising reverses this: $a^2 - b^2 = (a+b)(a-b)$.

$a^2-b^2=(a+b)(a-b)$

Minus sign only

The formula needs a minus between the squares. $a^2 + b^2$ won't work.

Square roots

$a$ and $b$ are the square roots of each term, not the original numbers.

Expand to verify

$(x+5)(x-5) = x^2 - 25$. Always check if unsure.

What You'll Master

objectives

Know

The difference of squares formula $a^2 - b^2 = (a+b)(a-b)$
Common perfect squares up to 100
Why the middle terms cancel on expansion

Understand

Why $a^2 + b^2$ cannot be factorised over the reals
Why you must take the square root, not halve the number
When to extract a common factor first

Can Do

Recognise a difference of squares at a glance
Factorise $a^2 - b^2$ where $a$ and $b$ are numbers, variables, or both
Combine HCF extraction with difference of squares

Words You Need

vocabulary

Difference of squaresAn expression of the form $a^2 - b^2$ that factors into $(a+b)(a-b)$.

Perfect squareA number or term that is the square of something: $9 = 3^2$, $x^2 = (x)^2$, $4x^2 = (2x)^2$.

Square rootThe value that, when squared, gives the original term. $\sqrt{25} = 5$, $\sqrt{x^2} = x$.

Factorise fullyKeep factorising until no further factorising is possible — including common factors.

Sum of squares$a^2 + b^2$ — cannot be factorised using real numbers. Do not confuse with difference.

HCFHighest Common Factor — always check for one before applying any formula.

Spot the Trap

heads-up

Wrong: "$x^2 - 9 = (x - 9)(x + 9)$" — forgot to take the square root of 9.

Right: $x^2 - 9 = (x + 3)(x - 3)$ because $\sqrt{9} = 3$, not 9.

Wrong: "$x^2 + 16 = (x + 4)(x - 4)$" — sum of squares, not difference.

Right: $x^2 + 16$ cannot be factorised over the reals. Only $a^2 - b^2$ works.

The Pattern

+5 XP

The difference of squares formula is one of the most useful shortcuts in algebra. It works because the outer and inner products cancel when you expand $(a+b)(a-b)$, leaving only $a^2 - b^2$.

Expand $(a+b)(a-b)$: First $= a^2$, Outer $= -ab$, Inner $= +ab$, Last $= -b^2$. The $-ab$ and $+ab$ cancel. Result: $a^2 - b^2$.

$(a+b)(a-b)$
= a² - b²

Memorise the form

$a^2 - b^2 = (a+b)(a-b)$. Both brackets identical except the sign.

Middle terms vanish

$-ab + ab = 0$. That's why the result has only two terms.

Order does not matter

$(a-b)(a+b)$ gives the same result. Same brackets, same answer.

Spot the Square

+5 XP

You can't use the formula unless both terms are perfect squares. Memorise the common ones and learn to spot squares hiding inside coefficients and variables.

Numbers: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$. Variables: $x^2 = (x)^2$, $x^4 = (x^2)^2$. Combined: $9x^2 = (3x)^2$, $16a^2 = (4a)^2$.

$4x^2=(2x)^2$
root 49 = 7

Coefficients too

$4x^2 = (2x)^2$. The coefficient 4 is also a perfect square.

Even powers

$x^4 = (x^2)^2$, $x^6 = (x^3)^2$. Any even exponent is a square.

Halving is wrong

$x^2 - 25$ uses 5, not 25/2 = 12.5. Take the square root.

HCF First

+5 XP

Before you reach for the difference of squares formula, always check for a common factor. Extract it first, then look at what's left inside the brackets. This is how you factorise fully.

$3x^2 - 27$ looks like a difference of squares, but both terms share a factor of 3. Pull out the 3 first: $3(x^2 - 9)$. Now $x^2 - 9$ is a difference of squares: $3(x+3)(x-3)$.

$3x^2-27$
= 3(x+3)(x-3)

HCF before formula

Always extract common factors first. The formula only applies to what's left.

Fully factorised

$3(x^2 - 9)$ is not fully factorised. Keep going until nothing else factors.

Expand to verify

$3(x+3)(x-3) = 3(x^2 - 9) = 3x^2 - 27$. Check every step.

Sums Stay Put

+5 XP

A sum of squares — $a^2 + b^2$ — cannot be factorised using real numbers. There is no pair of brackets that multiplies to give $a^2 + b^2$. Only differences split.

$(a+b)(a+b) = a^2 + 2ab + b^2$ — too many terms. $(a+b)(a-b) = a^2 - b^2$ — gives a minus, not a plus. No real factorisation exists for $a^2 + b^2$.

$a^2 + b^2$
no real factors

Plus means stop

If you see $a^2 + b^2$, the expression is already fully factorised over reals.

Don't force it

$(a+b)(a+b)$ gives three terms, not two. There is no valid pair.

Minus is the key

The formula ONLY works for $a^2 - b^2$. The sign matters.

Watch Me Solve It · 3 examples

Watch Me Solve It · Basic difference of squares

+15 XP per step

PROBLEM

Factorise $x^2 - 25$.

1

Recognise the pattern

$x^2 - 25$ is a difference of two squares

Two terms, minus sign, both are perfect squares.
2

Identify $a$ and $b$

$a = x$ and $b = 5$ (since $25 = 5^2$)

Take the square root of each term, not the original number.
3

Apply the formula

$x^2 - 25 = (x + 5)(x - 5)$

$a^2 - b^2 = (a+b)(a-b)$. Substitute $a=x$ and $b=5$.
4

Check by expanding

$(x + 5)(x - 5) = x^2 - 5x + 5x - 25 = x^2 - 25$ ✓

The middle terms cancel. We get back to the original.

Answer$(x + 5)(x - 5)$

Watch Me Solve It · Coefficients as squares

+15 XP per step

PROBLEM

Factorise $4x^2 - 49$.

1

Check for common factor

No common factor — proceed to formula
2

Identify $a$ and $b$

$4x^2 = (2x)^2$ so $a = 2x$

$49 = 7^2$ so $b = 7$

The coefficient 4 is also a perfect square. $\sqrt{4x^2} = 2x$.
3

Apply the formula

$4x^2 - 49 = (2x + 7)(2x - 7)$
4

Check by expanding

$(2x + 7)(2x - 7) = 4x^2 - 14x + 14x - 49 = 4x^2 - 49$ ✓

Answer$(2x + 7)(2x - 7)$

Watch Me Solve It · Common factor first

+15 XP per step

PROBLEM

Fully factorise $3x^2 - 27$.

1

Extract the HCF

$3x^2 - 27 = 3(x^2 - 9)$

HCF of $3x^2$ and $27$ is $3$. Always do this first.
2

Recognise difference of squares inside

$x^2 - 9 = x^2 - 3^2$

What's left in the bracket is now a difference of two perfect squares.
3

Apply the formula

$3(x^2 - 9) = 3(x + 3)(x - 3)$

$a = x$, $b = 3$. The 3 outside stays outside.
4

Check by expanding

$3(x + 3)(x - 3) = 3(x^2 - 9) = 3x^2 - 27$ ✓

Two-step factorising verified. Fully factorised.

Answer$3(x + 3)(x - 3)$

Common Pitfalls

heads-up

Forgetting to take the square root

$x^2 - 9 \rightarrow (x - 9)(x + 9)$ — used 9 instead of $\sqrt{9} = 3$.

Fix: always ask "what number, when squared, gives this term?"

Trying to factorise a sum of squares

$x^2 + 16 \rightarrow (x + 4)(x - 4)$ — this is a sum, not a difference. The result would be $x^2 - 16$, not $x^2 + 16$.

Fix: $x^2 + 16$ cannot be factorised over the reals. Leave it as is.

Missing the common factor

$2x^2 - 50 \rightarrow (x + 5)(x - 5)$ — forgot to extract the 2 first.

Fix: $2x^2 - 50 = 2(x^2 - 25) = 2(x + 5)(x - 5)$. HCF first, formula second.

Copy Into Your Books

The Formula

$a^2 - b^2 = (a+b)(a-b)$
Only works for differences
$a$ and $b$ are square roots

Perfect Squares

$1, 4, 9, 16, 25, 36, 49, 64, 81, 100$
$x^2 = (x)^2$, $4x^2 = (2x)^2$
Any even power is a square

Multi-step

Extract HCF first
Then apply the formula
Expand to verify

What Not to Do

$a^2 + b^2$ does not factorise
Don't halve — take the root
Don't stop at $3(x^2 - 9)$

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Mixed

4 problems

Four problems mixing basic difference of squares, coefficients, common factors and two variables. Work each one, then reveal the answer to check.

1 Factorise $x^2 - 81$.

$a = x$, $b = 9$ since $81 = 9^2$$= (x + 9)(x - 9)$
2 Factorise $16y^2 - 25$.

$16y^2 = (4y)^2$ and $25 = 5^2$$= (4y + 5)(4y - 5)$
3 Fully factorise $2x^2 - 50$.

HCF is 2: $2(x^2 - 25)$. Then $x^2 - 25 = (x+5)(x-5)$$= 2(x + 5)(x - 5)$
4 Factorise $9a^2 - 16b^2$.

$9a^2 = (3a)^2$ and $16b^2 = (4b)^2$$= (3a + 4b)(3a - 4b)$

Complete in your workbook.

Quick Check · 5 questions

Factorise $x^2 - 25$.

+10 XP

Which of the following is a difference of squares?

+10 XP

Factorise $4x^2 - 49$.

+10 XP

Factorise $9a^2 - 16b^2$.

+10 XP

Fully factorise $2x^2 - 50$.

+10 XP

Show Your Working · 3 questions

Show Your Working

7 marks total

Apply Easy 2 MARKS

Q6. Factorise each of the following expressions fully.

(a) $x^2 - 81$ (1 mark)

(b) $16y^2 - 25$ (1 mark)

Answer in your workbook.

Analyse Medium 4 MARKS

Q7. For each expression below, state whether it is a difference of squares. If it is, factorise it fully. If it is not, explain why.

(a) $x^2 + 4$ (1 mark)

(b) $25 - 4x^2$ (2 marks)

Answer in your workbook.

Evaluate Hard 4 MARKS

Q8. A square sports field has side length $(x + 8)$ metres. A smaller square warm-up area inside it has side length $(x - 8)$ metres.

(a) Write an expression for the area of the field not covered by the warm-up area. (2 marks)

(b) Factorise this expression fully using the difference of squares. (2 marks)

Answer in your workbook.

Comprehensive Answers

Quick Check

1. A — $x^2 - 25 = x^2 - 5^2 = (x+5)(x-5)$.

2. B — $x^2 - 9 = x^2 - 3^2$ is a difference of two perfect squares with a minus sign.

3. C — $4x^2 = (2x)^2$ and $49 = 7^2$. So $(2x+7)(2x-7)$.

4. D — $9a^2 = (3a)^2$ and $16b^2 = (4b)^2$. So $(3a+4b)(3a-4b)$.

5. B — First take out 2: $2(x^2 - 25)$. Then $x^2 - 25 = (x+5)(x-5)$. So $2(x+5)(x-5)$.

Show Your Working Model Answers

Q6 (2 marks): (a) $(x+9)(x-9)$ [1]. (b) $(4y+5)(4y-5)$ [1].

Q7 (4 marks): (a) Not a difference of squares — it is a sum, not a difference [1]. (b) $25 - 4x^2 = 5^2 - (2x)^2$ [1] $= (5+2x)(5-2x)$ [1]. (c) $3(x^2 - 4)$ [1] $= 3(x+2)(x-2)$ [1].

Q8 (4 marks): (a) $(x+8)^2 - (x-8)^2$ [1] $= x^2 + 16x + 64 - (x^2 - 16x + 64) = 32x$ m$^2$ [1]. (b) $[(x+8)+(x-8)][(x+8)-(x-8)]$ [1] $= 2x \cdot 16 = 32x$ m$^2$ [1].

Stretch Challenge · +25 XP, +10 coins

Factorise Twice

Fully factorise $x^4 - 16$. Hint: apply the difference of squares formula once, then look carefully at what remains inside each bracket — one of them might be factorisable again.

Reveal solution

$x^4 - 16 = (x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)$.

Now $x^2 - 4$ is also a difference of squares: $(x+2)(x-2)$.

But $x^2 + 4$ is a sum of squares — it cannot be factorised over the reals.

Fully factorised: $(x^2 + 4)(x + 2)(x - 2)$.

Quick Review

Formula

$a^2 - b^2 = (a+b)(a-b)$

Perfect Squares

Recognise $1,4,9,16,25,36,49,64,81,100$

Square Roots

$a$ and $b$ are the square roots of each term

Common Factor

Always take out HCF first

Sum of Squares

$a^2 + b^2$ cannot be factorised this way

Check

Expand to verify your answer

Interactive: Difference of Squares Explorer

Practise recognising and factorising difference of squares expressions with a visual area model. Adjust the terms and watch the factorisation unfold.

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