Year 8 - Algebra - Free practice

Year 8 Algebra Worksheets & Practice Tests

Expanding, factorising and simplifying linear expressions.

Quick answer

How do you simplify and expand algebraic expressions in Year 8?

Simplify by collecting like terms (same variable and power), expand brackets using the distributive law, and factorise by taking out the highest common factor. For example, 3(2x + 4) expands to 6x + 12, and 6x + 12 factorises back to 3(2x + 4).

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Six questions across Foundation, Standard and Extension, with full worked solutions, written to the Australian Curriculum.

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Worked examples

Year 8 algebra: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Collecting like terms and expanding brackets
Expand and simplify $3(2x + 5) - 4x + 7$ .
Show worked solution
Key idea: Use the distributive law to expand the bracket first, then collect like terms.
1. Expand the bracket: $3(2x + 5) = 6x + 15$
2. Rewrite the full expression: $6x + 15 - 4x + 7$
3. Collect the $x$ terms: $6x - 4x = 2x$
4. Collect the constant terms: $15 + 7 = 22$
5. Write the simplified answer: $2x + 22$
Answer: $2x + 22$
Example 2 · Factorising by taking out a common factor
Factorise fully $12x^2 - 8x$ .
Show worked solution
Key idea: Find the highest common factor (HCF) of all terms, write it outside the bracket, then divide each term by it.
1. Find the HCF of $12x^2$ and $8x$ : the HCF of 12 and 8 is 4, and both terms share a factor of $x$ , so HCF $= 4x$ .
2. Divide each term by $4x$ : $12x^2 \div 4x = 3x$ and $8x \div 4x = 2$ .
3. Write the factorised form: $4x(3x - 2)$ .
4. Check by expanding: $4x(3x - 2) = 12x^2 - 8x$ ✓
Answer: $4x(3x - 2)$
Example 3 · Simplifying and substituting in a real-life context
A school canteen sells pies for $p$ dollars each and drinks for $d$ dollars each. On Monday, Aisha buys 3 pies and 2 drinks. On Tuesday, she buys 1 pie and 4 drinks. Write a single simplified expression for her total spending over both days. Then find the total when $p = 4$ and $d = 2.50$ .
Show worked solution
Key idea: Write an expression for each day, add them, collect like terms, then substitute.
1. Monday's cost: $3p + 2d$
2. Tuesday's cost: $p + 4d$
3. Total: $(3p + 2d) + (p + 4d) = 3p + p + 2d + 4d$
4. Simplify: $4p + 6d$
5. Substitute $p = 4$ and $d = 2.50$ : $4(4) + 6(2.50) = 16 + 15 = 31$
Answer: The simplified expression is $4p + 6d$ ; the total is $31.

Common mistakes

Where students usually lose marks

Forgetting to multiply every term inside the bracket

$3(2x + 5) = 6x + 5$

The distributive law requires you to multiply every term inside the bracket by the number outside. Here, both $2x$ and $5$ must be multiplied by 3, giving $6x + 15$ .

Treating unlike terms as like terms

$5x + 3x^2 = 8x^3$

$5x$ and $3x^2$ have different powers of $x$ , so they are unlike terms and cannot be added together. Only terms with exactly the same variable and power can be combined.

Taking out a factor that is not the highest common factor

Factorising $12x^2 - 8x$ as $2(6x^2 - 4x)$

While 2 is a common factor, the expression is not fully factorised. Always find the highest common factor of both the coefficients and the variables. Here the HCF is $4x$ , giving $4x(3x - 2)$ .

Sign errors when expanding a bracket with a negative multiplier

$-2(x - 3) = -2x - 6$

When the multiplier is negative, multiplying a negative term inside the bracket gives a positive result. Here, $-2 \times (-3) = +6$ , so the correct expansion is $-2x + 6$ . Write out each multiplication separately if you are unsure.

Try one yourself

Your turn

Expand and simplify the expression $4(3x + 2) + 5x$ .

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About this worksheet

Why we built it

Year 8 algebra is where students move beyond writing expressions and start manipulating them - expanding brackets, factorising out common terms, collecting like terms across longer expressions. These worksheets give your student the repetition they need to make those moves automatic before quadratics arrive in Year 9.

What's covered

Sub-skills your student will practise

✓Collecting like terms with multiple variables
✓Expanding single-bracket expressions using the distributive law
✓Factorising by taking out a common factor
✓Rearranging expressions and applying inverse operations
✓Substituting values into simplified expressions

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Frequently asked

Questions parents ask about algebra

What does Year 8 algebra cover that Year 7 didn't?

Year 8 introduces expanding brackets using the distributive law, factorising by taking out a common factor, and rearranging expressions across both sides of an equation. Year 7 stayed mostly with writing and substituting; Year 8 is where students start manipulating.

How do students factorise an algebraic expression?

Find the largest common factor between the terms and pull it out. For 6x + 12, the common factor is 6, leaving 6(x + 2). For 4x^2 + 8x, the common factor is 4x, leaving 4x(x + 2).

What's the trickiest part of Year 8 algebra?

Keeping track of signs when expanding brackets that have negatives in front. -(x + 3) expands to -x - 3, not -x + 3. This is the single most common mistake on Year 8 tests.

How does Year 8 algebra prepare students for Year 9?

Year 9 introduces binomial expansion ((x + 2)(x + 3)) and factorising quadratics. Both build directly on the single-bracket distribution and common-factor work that Year 8 cements.

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