Mathematics • Year 7 • Unit 2 • Lesson 10

Factorising in the Real World

Use factorising — the reverse of expanding — to split an expression into "how many groups × what's in each group". A great tool for share-it-equally problems, recognising a hidden common factor, and rewriting a total in product form.

Apply · Real-World Maths

1. Word problems

Each problem uses ideas from Lesson 10: find the HCF of the terms, divide each term, and write as HCF × (...). Show your working — answers without working only get half marks.

1.1 — Identical party bags. Mia is making party bags. Across all the bags she has 4x chocolates and 8 stickers. She wants to put the same number of items into each bag, using up everything, with as many bags as possible.

(a) Factorise 4x + 8 to figure out how many bags she should make and what goes inside each bag.
(b) How many chocolates and stickers go into each bag?    2 marks

Stuck? 4x + 8 = 4(x + 2). The 4 outside is the number of bags. The (x + 2) is the contents of each bag.

1.2 — Rectangle dimensions. A rectangle has area 6x + 9 square centimetres. We want to find a sensible pair of "side lengths" (a width and a length) that could give that area.

(a) Factorise 6x + 9.
(b) Use your factorisation to suggest one possible pair of side lengths for the rectangle.    2 marks

Stuck? 6x + 9 = 3(2x + 3). So width = 3, length = (2x + 3) works.

1.3 — Sports day teams. Year 7 has x² students in one cohort and 7x students in another. The teachers want to factorise the total x² + 7x to see how to form equal teams.

(a) Factorise x² + 7x.
(b) Using your factorisation, describe what the "number of teams" is and what's "inside each team", in plain English.    3 marks

Stuck? x² + 7x = x(x + 7). So there are x teams, each containing (x + 7) students.

1.4 — School donation. A school receives 12x² in cash donations and 8x in food vouchers. The school wants to express the total receipt 12x² + 8x as "an amount per supporter × number of supporters".

(a) Fully factorise 12x² + 8x.
(b) Read your factorisation as "(amount per supporter) × (number of supporters)" — write down which factor is which and why your choice makes sense.    3 marks

Stuck? 12x² + 8x = 4x(3x + 2). One sensible reading: 4x supporters, each contributing (3x + 2). (Or vice versa — either reading is fine.)

1.5 — Painting walls. A painter calculates the total paint needed for two walls. Wall 1 needs 5a²b litres and Wall 2 needs 10ab² litres.

(a) Write the total as 5a²b + 10ab² and fully factorise it.
(b) Check your factorisation by expanding back.    3 marks

Stuck? Numbers HCF = 5. a in both terms (lowest power a). b in both terms (lowest power b). HCF = 5ab.

2. Explain your thinking

This question is about communication, not just symbols. Use full sentences. 4 marks

2.1 A classmate writes: "12x² + 8x = 2(6x² + 4x). Done." Use your own words to explain (i) why this is correct as far as it goes BUT not fully factorised, (ii) what "fully factorised" means, (iii) finish the job and write 12x² + 8x in its fully factorised form. Use a check (expand back to the original) to confirm.

Stuck? The 2 is a common factor — but it's not the HIGHEST one. The numbers share 4 as well. And both terms have an x — so x is in the HCF too. Overall HCF = 4x.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Identical party bags

(a) HCF(4, 8) = 4. So 4x + 8 = 4(x + 2). Mia can make 4 bags, each containing (x + 2) items.
(b) Each bag has x chocolates and 2 stickers.

1.2 — Rectangle dimensions

(a) HCF(6, 9) = 3. 6x + 9 = 3(2x + 3).
(b) A sensible pair of sides: width = 3 cm, length = (2x + 3) cm. Check: area = 3 × (2x + 3) = 6x + 9 ✓.

1.3 — Sports day teams

(a) HCF: numbers 1, variable x. HCF = x. x² + 7x = x(x + 7).
(b) Reading: there are x teams, and each team has (x + 7) students. (Or the other way round — both readings of HCF × (...) are valid in a real-world setting.)

1.4 — School donation

(a) HCF: numbers 4, variable x. HCF = 4x. 12x² + 8x = 4x(3x + 2).
(b) Sensible reading: there are 4x supporters, each contributing (3x + 2). Why this makes sense: the 4x captures both the size of the donor pool and a scaling factor x; the (3x + 2) is the per-donor contribution.

1.5 — Painting walls

(a) Total = 5a²b + 10ab². Numbers: HCF(5, 10) = 5. a in both (lowest power a). b in both (lowest power b). HCF = 5ab. Divide: 5a²b ÷ 5ab = a; 10ab² ÷ 5ab = 2b. Answer: 5ab(a + 2b) litres.
(b) Check by expanding: 5ab × a = 5a²b ✓, 5ab × 2b = 10ab² ✓.

2.1 — Explain your thinking (sample response)

(i) 12x² + 8x = 2(6x² + 4x) is technically correct — 2 IS a common factor, so when you take it out and expand back you get the original. But it's not fully factorised, because (6x² + 4x) still has its own common factor (you can still pull out 2x).
(ii) "Fully factorised" means the highest common factor has been taken out — so there is NO common factor left inside the brackets. Once you can't pull anything else out, you're done.
(iii) For 12x² + 8x, the HCF is actually 4x, not 2. Numbers: HCF(12, 8) = 4. Variable: x is in both terms (lowest power x¹). Divide: 12x² ÷ 4x = 3x; 8x ÷ 4x = 2. Fully factorised: 4x(3x + 2).
Check by expanding back: 4x × 3x = 12x² ✓ and 4x × 2 = 8x ✓ — matches the original.

Marking: 1 for explaining why 2(6x² + 4x) is partial, not wrong; 1 for clear definition of "fully factorised"; 1 for stating the correct full factorisation 4x(3x + 2); 1 for an explicit expansion check.