Algebraic expressions (expand & factorise)

Start here: two moves that undo each other

Expanding and factorising are opposite moves. Expanding multiplies a bracket out; factorising puts the bracket back.

\underbrace{3(x + 4)}_{\text{factorised}} \;\xrightarrow{\text{expand}}\; \underbrace{3x + 12}_{\text{expanded}} \;\xrightarrow{\text{factorise}}\; \underbrace{3(x + 4)}_{\text{back}}

If you can do one direction, you can do the other — you just need to know which move the question is asking for.

What you will learn

multiply an algebraic term through a bracket (expand),
collect like terms to simplify,
reverse expansion by pulling out the highest common factor (factorise),
simplify algebraic fractions of the form $\dfrac{\text{linear expression}}{\text{number}}$ ,
work with the associative, commutative, and distributive laws confidently.

1. Expand a single bracket

The distributive law: multiply every term inside the bracket by the factor in front.

Distributive law

a(b + c) = ab + ac, \qquad a(b - c) = ab - ac.

Worked example E Very easy: expanding a simple bracket

Expand $\;3(x + 4)$ .

Multiply the $3$ by every term inside the bracket: $3$ times $x$ is $3x$ , and $3$ times $4$ is $12$ .

3(x + 4) = 3x + 12.

Worked example 1 Expanding

Expand $\;4(2x - 3)$ .

4(2x - 3) = 4 \cdot 2x - 4 \cdot 3 = 8x - 12.

Worked example 2 Expanding a negative

Expand $\;-3(x - 5)$ .

-3(x - 5) = -3 \cdot x - (-3) \cdot 5 = -3x + 15.

2. Expand and collect like terms

Worked example 3 Two brackets, then simplify

Simplify $\;5(x + 2) - 3(x - 4)$ .

Expand: $5x + 10 - 3x + 12$ .
Collect like terms: $(5x - 3x) + (10 + 12) = 2x + 22$ .

Worked example 4 Variables on both terms

Simplify $\;2a(3a + 5) - a(a - 4)$ .

Expand: $6a^2 + 10a - a^2 + 4a$ .
Collect: $5a^2 + 14a$ .

3. Factorise by taking out the highest common factor (HCF)

Factorising is the reverse of expanding. Find the HCF of the terms and write the expression as (HCF) × (what’s left).

Factorising linear expressions

Single term factor

ab + ac = a(b + c).

Look for a factor common to every term - number, variable, or both.

Negative factors

-2x - 6 = -2(x + 3).

Pulling out a negative flips the sign of each term inside.

Worked example F Very easy: factorise 2x + 6

$2x + 6$ has no bracket. Is there something common to both terms? $2$ divides both ( $2x$ and $6 = 2 \times 3$ ). Pull the $2$ out:

2x + 6 = 2(x + 3).

Check by expanding: $2(x + 3) = 2x + 6$ . ✓

Worked example 5 Numeric HCF

Factorise $\;12x + 18$ .

HCF of $12$ and $18$ is $6$ :

12x + 18 = 6(2x + 3).

Check by expanding: $6(2x + 3) = 12x + 18$ . ✓

Worked example 6 Variable in the HCF

Factorise $\;5x^2 + 10x$ .

Both terms share $5$ and $x$ , so HCF is $5x$ :

5x^2 + 10x = 5x(x + 2).

Worked example 7 Three-term

Factorise $\;6a^2 - 9a + 3$ .

HCF of $6, 9, 3$ is $3$ :

6a^2 - 9a + 3 = 3(2a^2 - 3a + 1).

4. Simplifying algebraic fractions

When every term in the numerator shares a factor with the denominator, you can cancel.

Worked example 8 Divide each term

Simplify $\;\dfrac{12x + 18}{6}$ .

\dfrac{12x}{6} + \dfrac{18}{6} = 2x + 3.

(Alternatively: factorise top as $6(2x + 3)$ , cancel the $6$ .)

Practice: Year 8 core

Fluency

Expand

Expand $3(x + 4)$ .
Expand $5(y - 2)$ .
Expand $-2(a + 6)$ .
Expand $-4(x - 3)$ .
Expand $7(2p + 1)$ .
Expand $x(x + 5)$ .
Expand $2m(3m - 4)$ .
Expand $-a(a - 7)$ .

Fluency

Expand and collect

Simplify $2(x + 3) + 3(x + 1)$ .
Simplify $4(y - 2) - 2(y - 5)$ .
Simplify $3(2a - 1) + 5 - (a + 4)$ .
Simplify $5p + 2(p + 3) - 4$ .
Simplify $x(x + 2) - 3(x - 1)$ .

Fluency

Factorise

Factorise $6x + 9$ .
Factorise $10y - 15$ .
Factorise $12a + 16b$ .
Factorise $7x^2 + 14x$ .
Factorise $9ab - 12a$ .
Factorise $-3x + 9$ .
Factorise $2x^2 - 8x + 6$ .
Factorise $4mn + 6m^2 - 2m$ .

Fluency

Algebraic fractions

Simplify $\dfrac{8x + 12}{4}$ .
Simplify $\dfrac{15a - 10}{5}$ .
Simplify $\dfrac{6x^2 + 9x}{3}$ .
Simplify $\dfrac{20ab - 15b}{5b}$ .

Reasoning

Explain and spot the mistake

Jed writes $3(x - 2) = 3x - 2$ . Is Jed correct? If not, what is the error?
Mira factorises $6x + 4$ as $2(3x + 4)$ . Is this right? If not, give the correct factorisation.
Explain why $\dfrac{2x + 6}{2} = x + 3$ and not $x + 6$ .
Write two different expressions that both equal $6x + 12$ and demonstrate they are equal by expanding one of them.

Problem solving

Applications

A rectangle has length $x + 3$ cm and width $5$ cm. Write and simplify expressions for the perimeter and the area.
A taxi charges a flag-fall of $4 plus $2 per kilometre. For a $k$ -km trip, write an expression for the cost, and factorise it.
Five students each contribute $x toward a gift costing $42. Write and simplify an expression for the change each gets back, assuming the total change is shared equally.
Two rectangles have areas $6x + 12$ and $4x + 8$ . Factorise each; what does the result tell you about the shapes?

Challenge

Reasoning

Harder reasoning

Simplify $\;(x + 2)(x + 3) - x(x + 5)$ . (Hint: expand each product first.)
Factorise fully $\;4x^2 y + 8xy^2$ .
A rectangle has sides $a$ and $b$ . A second rectangle has sides $2a$ and $\dfrac{b}{2}$ . Show that the two rectangles have the same area.
Simplify $\dfrac{3(x - 2) + 2(x + 1)}{5}$ .

Answers

Answer key

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

Show the full answer key

Year 8 core - answers

Fluency

Expand

$3x + 12$
$5y - 10$
$-2a - 12$
$-4x + 12$
$14p + 7$
$x^2 + 5x$
$6m^2 - 8m$
$-a^2 + 7a$

Fluency

Expand and collect

$5x + 9$ . Method: $2x + 6 + 3x + 3$ .
$2y + 2$ . Method: $4y - 8 - 2y + 10$ .
$5a - 2$ . Method: $6a - 3 + 5 - a - 4$ .
$7p + 2$ . Method: $5p + 2p + 6 - 4$ .
$x^2 - x + 3$ . Method: $x^2 + 2x - 3x + 3$ .

Fluency

Factorise

$3(2x + 3)$
$5(2y - 3)$
$4(3a + 4b)$
$7x(x + 2)$
$3a(3b - 4)$
$-3(x - 3)$ or equivalently $3(3 - x)$
$2(x^2 - 4x + 3)$
$2m(2n + 3m - 1)$

Fluency

Algebraic fractions

$2x + 3$
$3a - 2$
$2x^2 + 3x$
$4a - 3$ . Method: divide each term by $5b$ .

Reasoning

Explain and spot the mistake

Wrong. The $3$ must multiply every term inside the bracket: $3(x - 2) = 3x - 6$ . Jed forgot to multiply the $-2$ .
Wrong. The HCF of $6$ and $4$ is $2$ . Correct: $2(3x + 2)$ , not $2(3x + 4)$ (which expands to $6x + 8$ ).
Both terms on top must be divided by $2$ : $\dfrac{2x}{2} + \dfrac{6}{2} = x + 3$ . Dividing only the $2x$ and leaving the $6$ unchanged is wrong.
Many answers. Example: $6x + 12 = 6(x + 2) = 3(2x + 4) = 2(3x + 6)$ . Expanding $6(x + 2)$ returns $6x + 12$ . ✓

Problem solving

Applications

Perimeter $= 2(x + 3) + 2(5) = 2x + 16$ . Area $= 5(x + 3) = 5x + 15$ .
Cost $= 4 + 2k = 2(2 + k)$ (factorised).
Total change $= 5x - 42$ . Each gets $\dfrac{5x - 42}{5}$ . (When this is not an integer, the “equally” is idealised.)
Factorisations: $6x + 12 = 6(x + 2)$ ; $4x + 8 = 4(x + 2)$ . Both rectangles share the side length $(x + 2)$ - i.e. they have a side in common.

Challenge - answers

Reasoning

Harder reasoning

$6$ . Method: $(x+2)(x+3) = x^2 + 5x + 6$ ; subtract $x(x+5) = x^2 + 5x$ ; $x^2 + 5x + 6 - x^2 - 5x = 6$ .
$4xy(x + 2y)$ .
First area $= ab$ . Second area $= 2a \times \dfrac{b}{2} = ab$ . Same.
$x - \dfrac{4}{5}$ . Method: top $= 3x - 6 + 2x + 2 = 5x - 4$ ; divide by $5$ .

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