Mathematics • Year 7 • Unit 2 • Lesson 10
Factorising — Mixed Challenge
Pull everything from Lesson 10 together: numeric HCFs, variable HCFs, mixed (number + variable) HCFs, fully-factorised checks, two-variable factorising, and an open-ended factor puzzle. Plus one classic Year 7 "looks factorised but isn't fully" mistake.
1. Mixed problems — choose the right idea
Each question uses a different part of Lesson 10. Find the HCF first, then divide each term. Show your working. 2 marks each
1.1 Find the HCF of each pair: (a) 12 and 18 (b) 8x and 12 (c) 6a and 9a².
1.2 Factorise each: (a) 3x + 12 (b) 10x − 15 (c) 7a + 21.
1.3 Factorise each (variable in the HCF): (a) x² + 5x (b) 4a² + 6a (c) 8x³ + 12x².
1.4 Fully factorise the two-variable expressions: (a) 6ab + 9a (b) 4x²y + 8xy² (c) 15a²b − 10ab².
1.5 Fully factorise 12x² + 8x. Then check your answer by expanding back to the original.
1.6 Which of these is FULLY factorised? Explain in one sentence why the others are not. (a) 2(3x + 6) (b) 3(2x + 3) (c) 6(x + 1.5) (d) 3(2x + 4).
2. Find the mistake
Another Year 7 student has tried to fully factorise 6a² + 9a. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then redo the working correctly. 3 marks
Student's working — for 6a² + 9a:
Line 1: HCF of 6 and 9 is 3.
Line 2: a² has power 2, a has power 1. HCF of a² and a is a² (take the highest power).
Line 3: Overall HCF = 3a².
Line 4: Divide: 6a² ÷ 3a² = 2, 9a ÷ 3a² = 3/a. So 6a² + 9a = 3a²(2 + 3/a).
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write the corrected working in full, including the corrected final factorised form.
Stuck? Rule: HCF of variables uses the LOWEST power that appears in every term — not the highest. (Otherwise, you'd end up with a fraction inside the brackets, which means you over-factorised.)3. Open-ended challenge — build your own expression
This question has more than one correct answer. Show one that works and explain. 4 marks
3.1 Find an algebraic expression (with two or more terms) that fully factorises to 3x(2x + 5). Then find a different expression that fully factorises to 4(2x + 3).
For each one, write down the expression (in expanded form) FIRST, then factorise it back to confirm it matches the target.
Bonus: Now write an expression that fully factorises to 5ab(a + 2b) — the answer will have TWO variables.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — HCFs
(a) HCF(12, 18) = 6. (b) HCF(8x, 12): numbers HCF(8, 12) = 4. Variable x is only in the first term, so x is NOT in the HCF. Answer: 4. (c) HCF(6a, 9a²): numbers HCF(6, 9) = 3. a in both, lowest power a. Answer: 3a.
1.2 — Number-only HCFs
(a) 3x + 12 = 3(x + 4). (b) 10x − 15 = 5(2x − 3). (c) 7a + 21 = 7(a + 3).
1.3 — Variable in the HCF
(a) x² + 5x = x(x + 5). (b) 4a² + 6a = 2a(2a + 3). (c) 8x³ + 12x² = 4x²(2x + 3).
1.4 — Two-variable expressions
(a) 6ab + 9a: HCF(6, 9) = 3; a in both (lowest power a); b only in first term. HCF = 3a. Answer: 3a(2b + 3).
(b) 4x²y + 8xy²: HCF(4, 8) = 4; x in both (lowest power x); y in both (lowest power y). HCF = 4xy. Answer: 4xy(x + 2y).
(c) 15a²b − 10ab²: HCF(15, 10) = 5; a in both (lowest a); b in both (lowest b). HCF = 5ab. Answer: 5ab(3a − 2b).
1.5 — 12x² + 8x with check
HCF: numbers 4, variable x. HCF = 4x. Divide: 12x² ÷ 4x = 3x; 8x ÷ 4x = 2. Answer: 4x(3x + 2). Check by expanding: 4x × 3x = 12x² ✓, 4x × 2 = 8x ✓. Sum: 12x² + 8x ✓.
1.6 — Which is fully factorised?
Only (b) 3(2x + 3) is fully factorised. Inside (2x + 3), there is no common factor — 2 and 3 share only 1.
(a) 2(3x + 6) is not fully factorised because (3x + 6) still has a common factor of 3. Full form: 6(x + 2).
(c) 6(x + 1.5) is invalid as a Year 7 factorisation because the inside has a fraction/decimal — we want integer coefficients inside. The fully-factorised integer form of 6x + 9 is 3(2x + 3).
(d) 3(2x + 4) is not fully factorised because (2x + 4) still has a common factor of 2. Full form: 6(x + 2).
2 — Find the mistake
(a) The mistake is on Line 2.
(b) When finding the HCF of variables, you use the LOWEST power, not the highest. a² has power 2, a has power 1, so the HCF is a¹ = a, not a². (If you take too much out, you end up with fractions inside the brackets — a clear sign you've gone too far.)
(c) Corrected Line 2: HCF of a² and a is a. Corrected Line 3: Overall HCF = 3a. Corrected Line 4: 6a² ÷ 3a = 2a; 9a ÷ 3a = 3. So 6a² + 9a = 3a(2a + 3).
Check: 3a × 2a = 6a² ✓, 3a × 3 = 9a ✓.
3 — Open-ended (sample solutions)
Target 1: 3x(2x + 5). Expand to find the matching expression: 3x × 2x + 3x × 5 = 6x² + 15x. Factorise back: HCF(6, 15) = 3; x in both (lowest power x); HCF = 3x. So 6x² + 15x = 3x(2x + 5) ✓.
Target 2: 4(2x + 3). Expand: 4 × 2x + 4 × 3 = 8x + 12. Factorise back: HCF(8, 12) = 4; no variable in HCF (only first term has x); HCF = 4. So 8x + 12 = 4(2x + 3) ✓.
Bonus — target 5ab(a + 2b): Expand: 5ab × a + 5ab × 2b = 5a²b + 10ab². Factorise back: HCF = 5ab. So 5a²b + 10ab² = 5ab(a + 2b) ✓.
Marking: 1 for the first matching expression 6x² + 15x with factorisation check; 1 for the second matching expression 8x + 12 with check; 1 for both expansion/factorisation directions being shown; 1 for the two-variable bonus.