Mathematics • Year 10 • Unit 2 • Lesson 6
Factorising Monic Quadratics — Mixed Challenge
Pull together every idea from Lesson 6: pair hunting, sign rules (same vs opposite), repeated brackets, and verification by expanding. Choose the right tool, spot a Year 10 student's slip, then design your own factorisable quadratic that fits a target structure.
1. Mixed problems — choose the right sign rule
Each question uses a different idea from Lesson 6. Decide which sign rule applies before you list pairs. Show your working. 3 marks each
1.1 Factorise x² + 11x + 24.
1.2 Factorise x² − 9x + 20.
1.3 Factorise x² + 3x − 18.
1.4 Factorise x² − 4x − 21.
1.5 Factorise x² − 14x + 49. (Look for a perfect square — both numbers equal.)
1.6 A rectangle has area x² + 13x + 40 m². Factorise the area and state both side lengths. Then evaluate the area when x = 2.
2. Find the mistake
Another Year 10 student has tried to factorise x² − 2x − 15. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the factorisation correctly. 3 marks
Student's working — factorise x² − 2x − 15:
Line 1: c = −15 → mixed signs.
Line 2: Pairs of −15: (1, −15), (−1, 15), (3, −5), (−3, 5).
Line 3: Sums: −14, 14, −2, 2. So the right pair is (3, −5) since 3 + (−5) = −2 ✓.
Line 4: Answer: (x + 3)(x + 5).
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected factorisation, and verify by expanding.
Stuck? Revisit lesson § "Check by Expanding" — does (x + 3)(x + 5) really expand to x² − 2x − 15?3. Open-ended challenge — design your own factorisable quadratic
This question has many valid answers. Show every step. 4 marks
3.1 Design a monic quadratic of the form x² + bx + c that satisfies all four of:
• it factorises over the integers,
• c is negative (so the brackets have opposite signs),
• the absolute value of b is at least 4,
• when x = 5 your expression evaluates to a positive number.
In your submission, include:
(i) Your chosen quadratic.
(ii) The factorisation, with the factor-pair search shown.
(iii) The check at x = 5 showing your value is positive.
(iv) A verification by expanding your brackets.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — x² + 11x + 24
Pairs of 24: (1, 24), (2, 12), (3, 8), (4, 6). Sums: 25, 14, 11 ✓, 10. Answer: (x + 3)(x + 8).
1.2 — x² − 9x + 20
c > 0, b < 0 → both negative. Negative pairs of 20: (−1, −20), (−2, −10), (−4, −5). Sums: −21, −12, −9 ✓. Answer: (x − 4)(x − 5).
1.3 — x² + 3x − 18
c < 0 → mixed signs. Pairs of −18 with sum +3: try (−3, 6): sum 3 ✓. Answer: (x − 3)(x + 6).
1.4 — x² − 4x − 21
c < 0 → mixed signs. b < 0 → larger absolute value is negative. Pairs of −21: try (3, −7): sum −4 ✓. Answer: (x + 3)(x − 7).
1.5 — x² − 14x + 49
c > 0, b < 0 → both negative. Negative pairs of 49: (−1, −49), (−7, −7). Sums: −50, −14 ✓. Answer: (x − 7)(x − 7) = (x − 7)².
1.6 — Rectangle area x² + 13x + 40
Pairs of 40 summing to 13: (5, 8) ✓. Factorised: (x + 5)(x + 8). Side lengths: (x + 5) m and (x + 8) m.
x = 2 → sides 7 m and 10 m, area = 70 m². Check: 4 + 26 + 40 = 70. ✓
2 — Find the mistake
(a) The mistake is on Line 4.
(b) Line 3 correctly identified the pair as (3, −5), but Line 4 then wrote (x + 3)(x + 5) — the second sign was changed from − to +. The brackets must use the exact numbers found in Line 3, so the second bracket should be (x − 5), not (x + 5).
(c) Corrected: x² − 2x − 15 = (x + 3)(x − 5). Verify: (x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15. ✓
3 — Open-ended challenge (sample solutions)
Many valid answers. Two starter examples that meet all four conditions:
Sample 1 — x² + 5x − 14
(i) Quadratic chosen with c = −14 (negative ✓), |b| = 5 ≥ 4 ✓.
(ii) Pairs of −14: (−2, 7), (2, −7), (−1, 14), (1, −14). Sums: 5 ✓, −5, 13, −13. Factorisation: (x − 2)(x + 7).
(iii) At x = 5: (3)(12) = 36 > 0 ✓.
(iv) Check: (x − 2)(x + 7) = x² + 7x − 2x − 14 = x² + 5x − 14. ✓
Sample 2 — x² − 6x − 16
(i) c = −16 ✓, |b| = 6 ≥ 4 ✓.
(ii) Pairs of −16 with sum −6: try (2, −8): sum −6 ✓. Factorisation: (x + 2)(x − 8).
(iii) At x = 5: (7)(−3) = −21. ✗ negative — must adjust.
Try x² − 4x − 21 = (x + 3)(x − 7) instead. At x = 5: (8)(−2) = −16. Still negative. Adjust again: x² + 4x − 21 = (x − 3)(x + 7). At x = 5: (2)(12) = 24 > 0 ✓.
Check: (x − 3)(x + 7) = x² + 7x − 3x − 21 = x² + 4x − 21. ✓
Marking: 1 for a valid quadratic meeting all four constraints; 1 for a clean factorisation with pair search; 1 for the positive value at x = 5; 1 for the FOIL verification. Full marks for any valid construction.