Year 8 Mathematics | Victorian Curriculum 2.0
Algebraic expressions (expand & factorise)
Topic 06 | Number & Algebra | Practice

Start here: two moves that undo each other

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

3(x+4)⏟factorised  →expand  3x+12⏟expanded  →factorise  3(x+4)⏟back\underbrace{3(x + 4)}_{\text{factorised}} \;\xrightarrow{\text{expand}}\; \underbrace{3x + 12}_{\text{expanded}} \;\xrightarrow{\text{factorise}}\; \underbrace{3(x + 4)}_{\text{back}}factorised3(x+4)​​expand​expanded3x+12​​factorise​back3(x+4)​​

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

Try mentally first

Read the question and ask “brackets in or brackets out?” If the expression has brackets, expanding takes them off. If it has none, factorising pulls a common factor out and puts brackets around what’s left.

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 linear expressionnumber\dfrac{\text{linear expression}}{\text{number}}numberlinear expression​,
  • work with the associative, commutative, and distributive laws confidently.
Where you'll see this
  • Physics formulas: simplifying ut+12at2ut + \tfrac{1}{2}at^2ut+21​at2 or factorising quantities.
  • Finance: compound cost calculations, where you pull common rates out.
  • Spreadsheets: rewriting cell formulas in a simpler equivalent form.
  • Area problems: writing one expression for a composite area and factorising it.

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,a(b−c)=ab−ac.a(b + c) = ab + ac, \qquad a(b - c) = ab - ac.a(b+c)=ab+ac,a(b−c)=ab−ac.
Worked example E Very easy: expanding a simple bracket

Expand   3(x+4)\;3(x + 4)3(x+4).

Multiply the 333 by every term inside the bracket: 333 times xxx is 3x3x3x, and 333 times 444 is 121212.

3(x+4)=3x+12.3(x + 4) = 3x + 12.3(x+4)=3x+12.
Worked example 1 Expanding

Expand   4(2x−3)\;4(2x - 3)4(2x−3).

4(2x−3)=4⋅2x−4⋅3=8x−12.4(2x - 3) = 4 \cdot 2x - 4 \cdot 3 = 8x - 12.4(2x−3)=4⋅2x−4⋅3=8x−12.
Worked example 2 Expanding a negative

Expand   −3(x−5)\;-3(x - 5)−3(x−5).

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

2. Expand and collect like terms

Worked example 3 Two brackets, then simplify

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

  1. Expand: 5x+10−3x+125x + 10 - 3x + 125x+10−3x+12.
  2. Collect like terms: (5x−3x)+(10+12)=2x+22(5x - 3x) + (10 + 12) = 2x + 22(5x−3x)+(10+12)=2x+22.
Worked example 4 Variables on both terms

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

  1. Expand: 6a2+10a−a2+4a6a^2 + 10a - a^2 + 4a6a2+10a−a2+4a.
  2. Collect: 5a2+14a5a^2 + 14a5a2+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).ab + ac = a(b + c).ab+ac=a(b+c).

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

Negative factors
−2x−6=−2(x+3).-2x - 6 = -2(x + 3).−2x−6=−2(x+3).

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

Common mistake: multiplying only the first term

3(x+4)3(x + 4)3(x+4) is not 3x+43x + 43x+4. The 333 has to touch every term inside the bracket. A quick check: expand your answer back out — if you don’t land on the original, you’ve made the mistake.

Worked example F Very easy: factorise 2x + 6

2x+62x + 62x+6 has no bracket. Is there something common to both terms? 222 divides both (2x2x2x and 6=2×36 = 2 \times 36=2×3). Pull the 222 out:

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

Check by expanding: 2(x+3)=2x+62(x + 3) = 2x + 62(x+3)=2x+6. ✓

Worked example 5 Numeric HCF

Factorise   12x+18\;12x + 1812x+18.

HCF of 121212 and 181818 is 666:

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

Check by expanding: 6(2x+3)=12x+186(2x + 3) = 12x + 186(2x+3)=12x+18. ✓

Worked example 6 Variable in the HCF

Factorise   5x2+10x\;5x^2 + 10x5x2+10x.

Both terms share 555 and xxx, so HCF is 5x5x5x:

5x2+10x=5x(x+2).5x^2 + 10x = 5x(x + 2).5x2+10x=5x(x+2).
Worked example 7 Three-term

Factorise   6a2−9a+3\;6a^2 - 9a + 36a2−9a+3.

HCF of 6,9,36, 9, 36,9,3 is 333:

6a2−9a+3=3(2a2−3a+1).6a^2 - 9a + 3 = 3(2a^2 - 3a + 1).6a2−9a+3=3(2a2−3a+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   12x+186\;\dfrac{12x + 18}{6}612x+18​.

12x6+186=2x+3.\dfrac{12x}{6} + \dfrac{18}{6} = 2x + 3.612x​+618​=2x+3.

(Alternatively: factorise top as 6(2x+3)6(2x + 3)6(2x+3), cancel the 666.)


Practice: Year 8 core

Fluency

Expand

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

Expand and collect

    1. Simplify 2(x+3)+3(x+1)2(x + 3) + 3(x + 1)2(x+3)+3(x+1).
    2. Simplify 4(y−2)−2(y−5)4(y - 2) - 2(y - 5)4(y−2)−2(y−5).
    3. Simplify 3(2a−1)+5−(a+4)3(2a - 1) + 5 - (a + 4)3(2a−1)+5−(a+4).
    4. Simplify 5p+2(p+3)−45p + 2(p + 3) - 45p+2(p+3)−4.
    5. Simplify x(x+2)−3(x−1)x(x + 2) - 3(x - 1)x(x+2)−3(x−1).
Fluency

Factorise

    1. Factorise 6x+96x + 96x+9.
    2. Factorise 10y−1510y - 1510y−15.
    3. Factorise 12a+16b12a + 16b12a+16b.
    4. Factorise 7x2+14x7x^2 + 14x7x2+14x.
    5. Factorise 9ab−12a9ab - 12a9ab−12a.
    6. Factorise −3x+9-3x + 9−3x+9.
    7. Factorise 2x2−8x+62x^2 - 8x + 62x2−8x+6.
    8. Factorise 4mn+6m2−2m4mn + 6m^2 - 2m4mn+6m2−2m.
Fluency

Algebraic fractions

    1. Simplify 8x+124\dfrac{8x + 12}{4}48x+12​.
    2. Simplify 15a−105\dfrac{15a - 10}{5}515a−10​.
    3. Simplify 6x2+9x3\dfrac{6x^2 + 9x}{3}36x2+9x​.
    4. Simplify 20ab−15b5b\dfrac{20ab - 15b}{5b}5b20ab−15b​.
Reasoning

Explain and spot the mistake

    1. Jed writes 3(x−2)=3x−23(x - 2) = 3x - 23(x−2)=3x−2. Is Jed correct? If not, what is the error?
    2. Mira factorises 6x+46x + 46x+4 as 2(3x+4)2(3x + 4)2(3x+4). Is this right? If not, give the correct factorisation.
    3. Explain why 2x+62=x+3\dfrac{2x + 6}{2} = x + 322x+6​=x+3 and not x+6x + 6x+6.
    4. Write two different expressions that both equal 6x+126x + 126x+12 and demonstrate they are equal by expanding one of them.
Problem solving

Applications

    1. A rectangle has length x+3x + 3x+3 cm and width 555 cm. Write and simplify expressions for the perimeter and the area.
    2. A taxi charges a flag-fall of $4 plus $2 per kilometre. For a kkk-km trip, write an expression for the cost, and factorise it.
    3. 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.
    4. Two rectangles have areas 6x+126x + 126x+12 and 4x+84x + 84x+8. Factorise each; what does the result tell you about the shapes?

Challenge

Reasoning

Harder reasoning

    1. Simplify   (x+2)(x+3)−x(x+5)\;(x + 2)(x + 3) - x(x + 5)(x+2)(x+3)−x(x+5). (Hint: expand each product first.)
    2. Factorise fully   4x2y+8xy2\;4x^2 y + 8xy^24x2y+8xy2.
    3. A rectangle has sides aaa and bbb. A second rectangle has sides 2a2a2a and b2\dfrac{b}{2}2b​. Show that the two rectangles have the same area.
    4. Simplify 3(x−2)+2(x+1)5\dfrac{3(x - 2) + 2(x + 1)}{5}53(x−2)+2(x+1)​.
Year 8 Mathematics study companion | Practice