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.)