Expanding and factorising - Year 9 Mathematics

What you will learn

expand a product of two binomials using the distributive law,
recognise and apply the perfect-square expansion patterns,
recognise and apply the difference-of-squares pattern,
factorise monic quadratic trinomials of the form $x^2 + bx + c$ ,
understand expanding and factorising as inverse operations.

Worked example 0 Real-world example: paving a courtyard

A courtyard is $(x + 3)$ metres long and $(x + 2)$ metres wide. What is its area in expanded form?

Area $= (x+3)(x+2)$ .
Expand: $x \times x + x \times 2 + 3 \times x + 3 \times 2$ .
Simplify: $x^2 + 2x + 3x + 6 = x^2 + 5x + 6$ .
If $x = 10$ , the courtyard is $13 \times 12 = 156$ m $^2$ , and $10^2 + 5(10) + 6 = 156$ . It checks.

Key idea: expanding a binomial product gives a quadratic expression. The area model below shows why each term appears.

Area model: (x + 3)(x + 2) = x² + 5x + 6. Each region represents one term in the expansion.

1. Expanding binomial products

To expand $(x + a)(x + b)$ , multiply each term in the first bracket by each term in the second:

General binomial product

(x + a)(x + b) = x^2 + (a + b)x + ab.

The middle coefficient is the sum of $a$ and $b$ ; the constant term is the product of $a$ and $b$ .

Worked example 1 Basic binomial expansion

Expand $(x + 4)(x + 5)$ .

Using the formula: $a = 4$ , $b = 5$ .

(x+4)(x+5) = x^2 + (4+5)x + (4)(5) = x^2 + 9x + 20.

Worked example 2 With a negative term

Expand $(x - 3)(x + 7)$ .

Here $a = -3$ , $b = 7$ .

(x-3)(x+7) = x^2 + (-3+7)x + (-3)(7) = x^2 + 4x - 21.

Worked example 3 Both terms negative

Expand $(x - 6)(x - 2)$ .

Here $a = -6$ , $b = -2$ .

(x-6)(x-2) = x^2 + (-6 + (-2))x + (-6)(-2) = x^2 - 8x + 12.

2. Special products

Two patterns appear so often that they deserve their own formulas.

Special binomial products

Perfect square (sum)

(x + a)^2 = x^2 + 2ax + a^2.

Perfect square (difference)

(x - a)^2 = x^2 - 2ax + a^2.

Difference of squares

(x + a)(x - a) = x^2 - a^2.

Worked example 4 Perfect square expansion

Expand $(x + 7)^2$ .

(x+7)^2 = x^2 + 2(7)x + 7^2 = x^2 + 14x + 49.

Worked example 5 Difference of squares

Expand $(x + 9)(x - 9)$ .

(x+9)(x-9) = x^2 - 9^2 = x^2 - 81.

There is no middle term — the $+9x$ and $-9x$ cancel exactly.

3. Factorising monic quadratics

Factorising reverses expanding. Given $x^2 + bx + c$ , you need two numbers that add to $b$ and multiply to $c$ .

Factorising x² + bx + c

x^2 + bx + c = (x + p)(x + q) \quad \text{where } p + q = b \text{ and } pq = c.

Worked example 6 Factorising with two positive numbers

Factorise $x^2 + 7x + 12$ .

Find two numbers that add to $7$ and multiply to $12$ :

Try $3$ and $4$ : $3 + 4 = 7$ and $3 \times 4 = 12$ . Yes.

$x^2 + 7x + 12 = (x + 3)(x + 4).$

Check by expanding: $(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ . Correct.

Worked example 7 Factorising with a negative constant

Factorise $x^2 + 2x - 15$ .

Need two numbers that add to $2$ and multiply to $-15$ . Since the product is negative, one number is positive and one is negative:

Try $5$ and $-3$ : $5 + (-3) = 2$ and $5 \times (-3) = -15$ . Yes.

$x^2 + 2x - 15 = (x + 5)(x - 3).$

Worked example 8 Factorising with a negative middle term

Factorise $x^2 - 9x + 20$ .

Need two numbers that add to $-9$ and multiply to $20$ . Both must be negative (negative sum, positive product):

Try $-4$ and $-5$ : $(-4) + (-5) = -9$ and $(-4)(-5) = 20$ . Yes.

$x^2 - 9x + 20 = (x - 4)(x - 5).$

4. Factorising special forms

The special products from Section 2 can be reversed:

Factorising special forms

Difference of squares

x^2 - a^2 = (x + a)(x - a).

Perfect square trinomial

x^2 + 2ax + a^2 = (x + a)^2, \qquad x^2 - 2ax + a^2 = (x - a)^2.

Worked example 9 Factorising a difference of squares

Factorise $x^2 - 49$ .

Recognise $49 = 7^2$ , so this is a difference of squares:

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

Worked example 10 Factorising a perfect square trinomial

Factorise $x^2 - 10x + 25$ .

Check: is $25 = 5^2$ and is $10 = 2 \times 5$ ? Yes. So:

$x^2 - 10x + 25 = (x - 5)^2.$

Practice

Fluency

Tier 1: basic skills

Expand $(x + 2)(x + 6)$ .
Expand $(x + 5)(x - 3)$ .
Expand $(x - 4)(x - 7)$ .
Expand $(x + 8)^2$ .
Expand $(x - 3)^2$ .
Expand $(x + 11)(x - 11)$ .
Factorise $x^2 + 8x + 15$ .
Factorise $x^2 - 5x + 6$ .
Factorise $x^2 - 36$ .
Factorise $x^2 + 12x + 36$ .

Reasoning

Tier 2: mixed practice

Expand and simplify $(x + 3)(x + 4) - (x + 1)(x + 2)$ .
Factorise $x^2 + x - 20$ .
Factorise $x^2 - 3x - 28$ .
A square garden has side $(x + 5)$ m. A path of width $1$ m surrounds it. Find the area of the path in expanded form.
Factorise $x^2 - 100$ and hence evaluate $998^2 - 1000^2$ mentally.
Show that $(x + a)^2 - (x - a)^2 = 4ax$ by expanding both sides.
Factorise $x^2 - 14x + 49$ .
Factorise $2x^2 + 14x + 24$ completely.

Reasoning

Tier 3: explain and apply

Explain why $(x + a)(x - a)$ has no $x$ term. Use the area model or algebra to justify.
A rectangle has area $x^2 + 9x + 18$ cm $^2$ . Find expressions for its length and width.
Without expanding, decide whether $(x + 3)^2$ and $x^2 + 9$ are equivalent. Explain.
Factorise $x^2 - 2x - 35$ and use your factorisation to solve $x^2 - 2x - 35 = 0$ .
Explain the connection between expanding and factorising using the analogy of multiplication and division of numbers.

Challenge

Reasoning

Harder reasoning

Expand $(2x + 3)(x - 4)$ . (Note: this is a non-monic product — the coefficient of $x^2$ is not $1$ .)
Factorise $x^4 - 16$ completely. (Hint: treat $x^4$ as $(x^2)^2$ and apply difference of squares twice.)
Prove that the sum of any two consecutive odd numbers is divisible by $4$ . (Hint: let the odd numbers be $2n - 1$ and $2n + 1$ , and use difference of squares.)
If $x + \dfrac{1}{x} = 5$ , find the value of $x^2 + \dfrac{1}{x^2}$ . (Hint: square both sides and simplify.)

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Tier 1

$x^2 + 8x + 12$ .
$x^2 + 2x - 15$ .
$x^2 - 11x + 28$ .
$x^2 + 16x + 64$ .
$x^2 - 6x + 9$ .
$x^2 - 121$ .
$(x + 3)(x + 5)$ . Method: $3 + 5 = 8$ and $3 \times 5 = 15$ .
$(x - 2)(x - 3)$ . Method: $(-2) + (-3) = -5$ and $(-2)(-3) = 6$ .
$(x + 6)(x - 6)$ . Difference of squares: $36 = 6^2$ .
$(x + 6)^2$ . Perfect square: $12 = 2 \times 6$ and $36 = 6^2$ .

Tier 2

$(x^2 + 7x + 12) - (x^2 + 3x + 2) = 4x + 10$ . The $x^2$ terms cancel.
$(x + 5)(x - 4)$ . Method: $5 + (-4) = 1$ and $5 \times (-4) = -20$ .
$(x - 7)(x + 4)$ . Method: $(-7) + 4 = -3$ and $(-7)(4) = -28$ .
Outer square side: $(x + 5 + 2) = (x + 7)$ m. Path area $= (x+7)^2 - (x+5)^2 = (x^2 + 14x + 49) - (x^2 + 10x + 25) = 4x + 24$ m $^2$ .
$x^2 - 100 = (x+10)(x-10)$ . So $998^2 - 1000^2 = (998+1000)(998-1000) = 1998 \times (-2) = -3996$ .
$(x+a)^2 = x^2 + 2ax + a^2$ . $(x-a)^2 = x^2 - 2ax + a^2$ . Subtracting: $(x^2 + 2ax + a^2) - (x^2 - 2ax + a^2) = 4ax$ .
$(x - 7)^2$ . Perfect square: $14 = 2 \times 7$ and $49 = 7^2$ .
$2(x^2 + 7x + 12) = 2(x + 3)(x + 4)$ . Factor out $2$ first, then factorise the trinomial.

Tier 3

When expanding $(x+a)(x-a)$ , the middle terms are $+ax$ and $-ax$ , which sum to zero. Using the area model: the two rectangular strips ( $a \times x$ and $x \times a$ ) have opposite signs and cancel, leaving only $x^2 - a^2$ .
$x^2 + 9x + 18 = (x + 3)(x + 6)$ . Method: $3 + 6 = 9$ and $3 \times 6 = 18$ . So the length is $(x + 6)$ cm and the width is $(x + 3)$ cm (or vice versa).
They are not equivalent. $(x+3)^2 = x^2 + 6x + 9$ , which has a $6x$ term that $x^2 + 9$ lacks. For example, at $x = 1$ : $(1+3)^2 = 16$ but $1 + 9 = 10$ .
$x^2 - 2x - 35 = (x - 7)(x + 5) = 0$ . So $x = 7$ or $x = -5$ .
Expanding is like multiplication: $3 \times 4 = 12$ breaks a product into a single value. Factorising is like finding factors: $12 = 3 \times 4$ rewrites a value as a product. They undo each other. In algebra, expanding turns $(x+3)(x+4)$ into $x^2 + 7x + 12$ , and factorising reverses the process.

Challenge

$(2x+3)(x-4) = 2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12$ .
$x^4 - 16 = (x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)$ . Then $x^2 - 4 = (x+2)(x-2)$ , so $x^4 - 16 = (x^2 + 4)(x + 2)(x - 2)$ . The factor $(x^2 + 4)$ does not factorise further over the reals.
Product of two consecutive odd numbers: $(2n-1)(2n+1) = 4n^2 - 1$ . Their sum is $(2n-1) + (2n+1) = 4n$ , which is divisible by $4$ . (Note: the question asks about the sum, not the product.)
Square both sides of $x + \dfrac{1}{x} = 5$ : $x^2 + 2 + \dfrac{1}{x^2} = 25$ . So $x^2 + \dfrac{1}{x^2} = 25 - 2 = 23$ .

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