Mathematics • Year 10 • Unit 2 • Lesson 6
Monic Quadratics in the Real World
Use factorisation to recover dimensions from rectangular areas, design garden beds, find zeros of paths, and check claims about factorisable expressions. Then explain (in your own words) the sign-rule decision that controls every factorisation.
1. Word problems
Each problem uses factorising x² + bx + c. Show your factor-pair search — a final bracket with no working only earns half marks.
1.1 — Rectangular garden bed. A garden bed has area A = x² + 7x + 10 square metres.
(a) Factorise the area expression.
(b) Write down two possible side lengths of the bed in terms of x.
(c) If x = 4, what are the actual side lengths and the area in m²? 4 marks
1.2 — Picture frame. A square photo (side x cm) is surrounded by a uniform border so the total area of photo plus border is x² + 8x + 16 cm².
(a) Factorise x² + 8x + 16 (look for a perfect square — both numbers equal).
(b) State the total outer side length in terms of x.
(c) How wide is the border (compared to the photo side x)? 3 marks
1.3 — Path width problem. A council planner writes that the cross-sectional area of a footpath is A = x² − 3x − 10 m², where x is a positive design variable in metres.
(a) Factorise A.
(b) For the area to be zero (i.e. the path just disappears), what value of x is required? You will get two candidates — only the positive one makes physical sense.
(c) Use your factorisation to evaluate A when x = 7. 4 marks
1.4 — Pricing model. A small business models its weekly profit (in $100s) as P = x² − 11x + 30, where x is the price set ($ per item).
(a) Factorise P.
(b) At which two prices is P = 0 (break-even)?
(c) What is P (in $100s) when x = 4? Is the business making profit or loss at that price? 4 marks
1.5 — Check the claim. A classmate writes: "x² + 6x + 9 = (x + 3)(x + 3) and x² − 6x + 9 = (x − 3)(x − 3)."
(a) Verify both factorisations by expanding.
(b) Why do both expressions factorise to a "repeated bracket"? (Hint: what is the relationship between b and the factor pair?)
(c) What value of c would make x² + 10x + c a repeated-bracket factorisation? 3 marks
2. Explain your thinking
This question is about communication, not just numbers. Use full sentences. 4 marks
2.1 A friend says: "When I factorise x² + bx + c, I never bother checking which sign goes where. I just write (x + something)(x + something else) and move on." Using everything from Lesson 6, explain (i) what error this habit will cause when c is positive and b is negative, (ii) what error it will cause when c is negative, and (iii) the one habit your friend should adopt to stop making sign mistakes. Use the words "same sign" and "opposite signs" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Garden bed area
(a) Pairs of 10: (1, 10), (2, 5). Sums: 11, 7 ✓. Factorised: (x + 2)(x + 5).
(b) Side lengths are (x + 2) m and (x + 5) m.
(c) x = 4 → sides are 6 m and 9 m. Area = 6 × 9 = 54 m². Check: 4² + 7(4) + 10 = 16 + 28 + 10 = 54. ✓
1.2 — Picture frame
(a) Pairs of 16: (1, 16), (2, 8), (4, 4). Sums: 17, 10, 8 ✓. Factorised: (x + 4)(x + 4) = (x + 4)².
(b) Total outer side length is (x + 4) cm.
(c) Outer side − inner side = (x + 4) − x = 4 cm total across both sides → border is 2 cm wide.
1.3 — Footpath area
(a) c < 0 → mixed signs. Pairs of −10 with sum −3: try (−5, 2): sum −3 ✓. Factorised: (x − 5)(x + 2).
(b) A = 0 when x − 5 = 0 or x + 2 = 0, so x = 5 or x = −2. The positive answer is x = 5 m.
(c) x = 7 → A = (7 − 5)(7 + 2) = 2 × 9 = 18 m².
1.4 — Pricing model
(a) c > 0 and b < 0 → both negative. Negative pairs of 30: (−1, −30), (−2, −15), (−3, −10), (−5, −6). Sums: −31, −17, −13, −11 ✓. Factorised: (x − 5)(x − 6).
(b) Break-even at x = $5 and x = $6.
(c) x = 4 → P = (4 − 5)(4 − 6) = (−1)(−2) = +2 (i.e. +$200). Profit. Check: 16 − 44 + 30 = 2. ✓
1.5 — Repeated brackets
(a) (x + 3)(x + 3) = x² + 3x + 3x + 9 = x² + 6x + 9 ✓. (x − 3)(x − 3) = x² − 3x − 3x + 9 = x² − 6x + 9 ✓.
(b) Both factorise to a repeated bracket because m = n: the same number multiplies to c AND adds (twice) to b. In each case 3 × 3 = 9 = c and 3 + 3 = 6 = |b|.
(c) For x² + 10x + c to be a repeated bracket, we need m + m = 10, so m = 5, and m × m = 25. So c = 25 → (x + 5)².
2.1 — Explain your thinking (sample response)
(i) When c is positive and b is negative, both numbers must be negative (same sign as b). The friend's habit of always writing (x + something)(x + something) will give the wrong signs — for example x² − 7x + 12 should be (x − 3)(x − 4), not (x + 3)(x + 4). (ii) When c is negative, the two numbers have opposite signs. Writing both as + (or both as −) cannot produce a negative product, so the factorisation simply cannot work. (iii) The habit to adopt is: look at the sign of c first. If c > 0, both numbers share the same sign as b. If c < 0, the numbers have opposite signs and the one with the larger absolute value carries b's sign. Then always check by expanding.
Marking: 1 for the c > 0, b < 0 error; 1 for the c < 0 impossibility; 1 for the "check c first" habit; 1 for correct use of both required phrases.