Mathematics • Year 10 • Unit 2 • Lesson 6
Factorising Monic Quadratics — Skill Drill
Build fluency with the pair-hunt method from Lesson 6: list factor pairs of c, check which pair adds to b, then apply the sign rules. One fully-worked example, one guided trace with blanks, then eight independent problems graded from foundation to extension.
1. I do — fully worked example
Factorise a monic quadratic step-by-step. Each line has a reason underneath so you can see why, not just what.
Problem. Factorise x² + 9x + 18.
Step 1 — Identify b and c.
b = 9, c = 18
Reason: we need two numbers that multiply to 18 and add to 9.
Step 2 — List factor pairs of 18 from 1 upward.
(1, 18), (2, 9), (3, 6)
Reason: systematic list means we cannot miss the right pair. Stop near √18 ≈ 4.2.
Step 3 — Check each sum against b = 9.
1 + 18 = 19, 2 + 9 = 11, 3 + 6 = 9 ✓
Reason: only (3, 6) hits the target. c > 0 and b > 0, so both numbers are positive.
Step 4 — Write the factors and check by expanding.
x² + 9x + 18 = (x + 3)(x + 6)
Check (FOIL): x² + 6x + 3x + 18 = x² + 9x + 18. ✓
Answer: (x + 3)(x + 6).
2. We do — fill in the missing steps
Same method as Section 1, but with the working faded. Fill in each blank. 5 marks
Problem. Factorise x² − 8x + 15.
Step 1 — Identify b and c: b = ____ and c = ____.
Step 2 — Sign decision. c is ___________ (positive / negative). So both numbers have the ___________ sign. b is negative, so both numbers are ___________.
Step 3 — List the negative factor pairs of 15:
(−1, −15) and (____, ____)
Step 4 — Check the sums:
(−1) + (−15) = ____ , (____) + (____) = −8 ✓
Step 5 — Write the answer:
x² − 8x + 15 = (x ____ )(x ____ )
3. You do — independent practice
Show your factor-pair search and your answer. The first four are foundation (positive c, positive b). The middle two are standard (sign variations). The last two are extension.
Foundation — both positive
3.1 Factorise x² + 7x + 12. 1 mark
3.2 Factorise x² + 8x + 12. 1 mark
3.3 Factorise x² + 5x + 6. 1 mark
3.4 Factorise x² + 10x + 21. 1 mark
Standard — sign variations
3.5 Factorise x² − 6x + 8. (c positive, b negative — both numbers negative.) 2 marks
3.6 Factorise x² + 2x − 8. (c negative — mixed signs; larger absolute value takes b's sign.) 2 marks
Extension — push your thinking
3.7 Factorise x² − 5x − 14. Show the full factor-pair list and the sum check. 3 marks
3.8 A student claims that x² + 5x + 9 = (x + 3)(x + 3). Without expanding, explain in one sentence how you can tell the claim is wrong from the factor pairs of 9. Then state whether x² + 5x + 9 can be factorised over the integers at all. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded x² − 8x + 15)
Step 1: b = −8, c = 15.
Step 2: c is positive. Both numbers have the same sign. b is negative, so both numbers are negative.
Step 3: Negative factor pairs of 15 are (−1, −15) and (−3, −5).
Step 4: (−1) + (−15) = −16. (−3) + (−5) = −8 ✓.
Step 5: x² − 8x + 15 = (x − 3)(x − 5).
3.1 — x² + 7x + 12
Pairs of 12: (1, 12), (2, 6), (3, 4). Sums: 13, 8, 7 ✓. Answer: (x + 3)(x + 4).
3.2 — x² + 8x + 12
Pairs of 12: (1, 12), (2, 6), (3, 4). Sums: 13, 8 ✓, 7. Answer: (x + 2)(x + 6).
3.3 — x² + 5x + 6
Pairs of 6: (1, 6), (2, 3). Sums: 7, 5 ✓. Answer: (x + 2)(x + 3).
3.4 — x² + 10x + 21
Pairs of 21: (1, 21), (3, 7). Sums: 22, 10 ✓. Answer: (x + 3)(x + 7).
3.5 — x² − 6x + 8
c > 0 and b < 0 → both negative. Negative pairs of 8: (−1, −8), (−2, −4). Sums: −9, −6 ✓. Answer: (x − 2)(x − 4).
3.6 — x² + 2x − 8
c < 0 → mixed signs. Mixed pairs of −8: (−1, 8), (1, −8), (−2, 4), (2, −4). Sums: 7, −7, 2 ✓, −2. Answer: (x − 2)(x + 4). Check: x² + 4x − 2x − 8 = x² + 2x − 8.
3.7 — x² − 5x − 14
c < 0 → mixed signs. Pairs of −14: (−1, 14), (1, −14), (−2, 7), (2, −7). Sums: 13, −13, 5, −5 ✓.
b is negative, so the larger absolute value (7) is negative.
Answer: (x + 2)(x − 7). Check: x² − 7x + 2x − 14 = x² − 5x − 14.
3.8 — Why (x + 3)(x + 3) is wrong for x² + 5x + 9
The factor pairs of 9 are (1, 9) and (3, 3). Their sums are 10 and 6 — neither equals 5, so no integer pair fits and the claim of (x + 3)(x + 3) is impossible. The quadratic cannot be factorised over the integers.
(x + 3)(x + 3) actually expands to x² + 6x + 9, not x² + 5x + 9.)