Factorising Quadratics
SPEC A4 • ALGEBRA — FACTORISING QUADRATICS ⭐
Two techniques: the product-and-sum method for monic quadratics, and the AC method for non-monic. Fluency here unlocks solving quadratics, sketching parabolas, and simplifying algebraic fractions.
What You'll Learn
A quadratic is an expression of the form ax² + bx + c. This lesson splits naturally into two parts based on the value of a.
Part 1 — Monic quadratics (F+H)
Monic: a = 1. Form: x² + bx + c
Method: find two numbers with product c and sum b
Sign analysis — positive/negative combinations
Part 2 — Non-monic quadratics HIGHER
Non-monic: a ≠ 1. Form: ax² + bx + c
AC method: split the middle term using two numbers with product ac and sum b
Authentic AQA question: Jun 2022 Paper 2 Q23
Why it matters ⭐
This is a high-priority topic. Lessons 2.05 (algebraic fractions), 2.21 (solving quadratics), 2.22 (completing the square) and 2.13 (quadratic graphs) all depend on being able to factorise fluently.
Part 1 Monic Quadratics
Monic
A quadratic is monic when the coefficient of x² is 1 (often written without a visible coefficient).x² + bx + c
MONIC ✓
x² + 7x + 12
x² − 5x + 6
p² − 15p + 14
Coefficient of x² (or p²) is 1.
NON-MONIC ✗
2x² + 7x + 3
3x² − 16x − 12
5x² + 13x − 6
Coefficient of x² is not 1. Harder — Higher only.
Monic Method: Product & Sum
The Rule
For x² + bx + c, find two numbers that:multiply to give c AND add to give bBoth conditions must hold simultaneously. Those two numbers become the constants in the brackets.
Why it works
If x² + bx + c = (x + p)(x + q), expand the right side:(x + p)(x + q) = x² + qx + px + pq = x² + (p + q)x + pqComparing: p + q = b (sum) and pq = c (product). Find p and q and you've factorised.
Sign Analysis — Four Cases
Before hunting for numbers, use the signs of b and c to narrow down the search.
Sign of bSign of cBoth factorsExample
++both positivex² + 7x + 12 → (x + 3)(x + 4)
−+both negativex² − 5x + 6 → (x − 2)(x − 3)
+−mixed; positive one is biggerx² + 2x − 15 → (x + 5)(x − 3)
−−mixed; negative one is biggerx² − 5x − 14 → (x − 7)(x + 2)
Shortcut logic
c positive means both factors have the same sign (their product is positive). c negative means the factors have different signs. Then use the sign of b to decide which is larger.
01Worked Example — Both Positive
Factorise: x² + 7x + 12 [2 marks]
Step 1 — Identify what we need
Product = 12 AND Sum = 7
Step 2 — List factor pairs of 12 (both positive, since c > 0 and b > 0)
1 × 12 (sum 13) ✗
2 × 6 (sum 8) ✗
3 × 4 (sum 7) ✓
Step 3 — Write the factorisation
x² + 7x + 12 = (x + 3)(x + 4)
Step 4 — Check by expanding
(x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✓
02Worked Example — Negative b, Positive c
Factorise: x² − 5x + 6 [2 marks]
Step 1 — Sign analysis
c = +6 (positive) and b = −5 (negative). From the sign table: both factors are negative.
Step 2 — List negative factor pairs of 6
(−1) × (−6) (sum −7) ✗
(−2) × (−3) (sum −5) ✓
Step 3 — Write the factorisation
x² − 5x + 6 = (x − 2)(x − 3)
Check
(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6 ✓
03Worked Example — Mixed Signs
Factorise: x² + 2x − 15 [2 marks]
Step 1 — Sign analysis
c = −15 (negative) → factors have opposite signs. b = +2 (positive) → the positive factor is larger in size.
Step 2 — List mixed-sign factor pairs of −15
1 × (−15) (sum −14) ✗
3 × (−5) (sum −2) ✗
5 × (−3) (sum +2) ✓
15 × (−1) (sum +14) ✗
Step 3 — Write the factorisation
x² + 2x − 15 = (x + 5)(x − 3)
Sign slip trap
Students write (x − 5)(x + 3) instead — swapping the signs. Check by expanding: (x − 5)(x + 3) = x² − 2x − 15, which has the wrong sign on the x term. Always verify.
MCE Worked Example — Monic
Factorise:
p² − 15p + 14
[2 marks]
Method
c = +14, b = −15. Both factors negative. Find two negative numbers with product 14 and sum −15.
Calculate
Pairs of 14: (−1)(−14) sum −15 ✓ or (−2)(−7) sum −9 ✗
Winner: −1 and −14
Evaluate
p² − 15p + 14
= (p − 1)(p − 14)
Mark scheme (typical monic 2-mark)
B2 for fully correct (p − 1)(p − 14). B1 for (p + a)(p + b) where either ab = 14 (correct product) or a + b = −15 (correct sum) but not both. Brackets in either order — both score.
Part 2 Non-Monic Quadratics HIGHER
When a ≠ 1, the product-and-sum shortcut no longer works directly. The reliable method is the AC method (also called "splitting the middle term").
Why monic trick fails for ax² + bx + c
If 2x² + 7x + 3 = (2x + p)(x + q), expanding gives:2x² + 2qx + px + pq = 2x² + (2q + p)x + pqNow the coefficient of x is 2q + p, not p + q. The leading coefficient "a" distorts the sum — we need a different approach.
The AC method in one sentence
Find two numbers with product a × c and sum b. Then split the middle term using those numbers, and factorise by grouping.
AC Method — Step by Step H
Step 0 — ALWAYS check for a common factor first
If all three coefficients a, b, c share a common factor, pull it out before applying AC. Example: 2x² + 10x + 12 = 2(x² + 5x + 6) = 2(x + 2)(x + 3). Missing this step means you write a messier non-monic expression when a simpler monic lurks inside.
The four steps
Compute ac — multiply the leading coefficient by the constant term.
Find two numbers with product ac and sum b.
Split the middle term using those two numbers (order doesn't matter).
Factorise by grouping — pair up the four terms and pull out common factors.
Worked walk-through: 2x² + 7x + 3
ac = 2 × 3 = 6. Need product 6, sum 7. → 6 and 1.Split: 2x² + 6x + 1x + 3Group: (2x² + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3)Factor: (2x + 1)(x + 3)
The grouping check
After splitting and grouping, both pairs must produce the same bracket. Here (x + 3) appears in both — confirm this, then pull it out as the shared factor.
04Worked Example — Non-Monic H
Factorise: 3x² − 11x − 4 [2 marks]
Step 1 — Compute ac
a = 3, c = −4 → ac = 3 × (−4) = −12
Step 2 — Find numbers: product −12, sum −11
Since product is negative, one number is negative. Try pairs:
(−12) × 1 = −12 ✓, sum = −11 ✓
Step 3 — Split and group
3x² − 12x + x − 4
= 3x(x − 4) + 1(x − 4)
Step 4 — Pull out the common bracket
= (3x + 1)(x − 4)
Check by expanding
(3x + 1)(x − 4) = 3x² − 12x + x − 4 = 3x² − 11x − 4 ✓
MCE Worked Example — Authentic AQA H
Factorise:
3x² − 16x − 12
[2 marks] JUN 2022 • P2 • Q23
Method
AC method. Compute ac = 3 × (−12) = −36. Find two numbers with product −36 and sum −16.
Calculate
Pairs for −36: (−18)(2) — sum −16 ✓
Split: 3x² − 18x + 2x − 12
Group: 3x(x − 6) + 2(x − 6)
Evaluate
= (3x + 2)(x − 6)
Real AQA mark scheme (Jun 2022)
B2 for the fully correct (3x + 2)(x − 6). B1 for (3x + a)(x + b) where ab = −12 or a + 3b = −16. The mark scheme condones brackets in either order and condones missing closing brackets. Common B1 responses include (3x + 6)(x − 2), (3x + 4)(x − 3), (3x + 12)(x − 1).
Always Check by Expanding
Every factorisation can be verified by expanding your answer and comparing to the original. It takes 15 seconds and catches sign errors that would cost 2 marks.
WEAK STUDENT ✗
Factorises x² − 5x + 6 as (x + 2)(x + 3) and writes the answer.
Didn't expand to verify. Expansion gives x² + 5x + 6 — wrong sign on the middle term. Both B-marks lost.
STRONG STUDENT ✓
Writes (x − 2)(x − 3), then expands as a check.
= x² − 3x − 2x + 6 = x² − 5x + 6 ✓
Matches. Confident to move on.
Why this earns partial credit too
Even if your factorised answer has a sign slip, it can still earn B1. The AQA rule: for a quadratic x² + bx + c, brackets (x + p)(x + q) score B1 if either the product pq = c or the sum p + q = b (just one condition). So a written factorisation attempt — even an imperfect one — banks at least one mark.
Examiner Alerts
Alert 1 — Checking sum only, not product
Students write: for x² + 7x + 10, they write (x + 4)(x + 3) — because 4 + 3 = 7.
Write instead: check the product too. 4 × 3 = 12, not 10. The correct pair is 5 and 2 (sum 7 AND product 10). Answer: (x + 5)(x + 2). Both conditions must hold.
Alert 2 — Sign confusion on negatives
Students write: for x² − 5x − 14, they try −7 and +2 incorrectly as (x − 7)(x − 2).
Write instead: if the pair is −7 and +2, the brackets are (x − 7)(x + 2) — matching the signs of the numbers exactly. (x − 7)(x − 2) = x² − 9x + 14, wrong answer.
Alert 3 — Skipping the check
Students write: an AC-method answer straight to the answer line without expanding to verify.
Write instead: expand your factorised form and compare to the original. Even if the final answer has a sign slip, showing the split-and-group working banks B1 partial credit under the AQA "correct product or correct sum" rule.
Common Errors
Students often writeCorrect versionWhy it matters
x² + 7x + 10 = (x + 4)(x + 3)= (x + 5)(x + 2)Checked sum only; product 4×3=12, not 10
x² − 5x + 6 = (x + 2)(x + 3)= (x − 2)(x − 3)Both factors should be negative (c positive, b negative)
x² − 5x − 14 = (x − 7)(x − 2)= (x − 7)(x + 2)c negative → opposite signs in brackets
3x² − 16x − 12 = (3x + 6)(x − 2)= (3x + 2)(x − 6)Expands to 3x² − 12 (middle terms cancel). Scores B1 only — correct product, wrong sum
Didn't expand to check before writing the answerAlways verify: (bracket)(bracket) = originalFree insurance — catches sign slips
Practice Question 1
Factorise:
x² + 4x − 21
[2 marks]
M
Method. c = −21 (negative) → opposite signs. b = +4 (positive) → the positive factor is larger. Find two numbers with product −21 and sum +4.
C
Calculate.
Pairs for −21: 1 × (−21) sum −20 ✗; 3 × (−7) sum −4 ✗; 7 × (−3) sum +4 ✓
Winner: 7 and −3.
E
Evaluate. Final answer: (x + 7)(x − 3).
Check: (x + 7)(x − 3) = x² − 3x + 7x − 21 = x² + 4x − 21 ✓
Practice Question 2 H
Factorise:
2x² + 11x + 12
[2 marks]
M
Method. AC method. Compute ac = 2 × 12 = 24. Find two numbers with product 24 and sum 11.
C
Calculate.
Pairs of 24: 1×24 (sum 25) ✗, 2×12 (sum 14) ✗, 3×8 (sum 11) ✓, 4×6 (sum 10) ✗
Split: 2x² + 3x + 8x + 12
Group: x(2x + 3) + 4(2x + 3)
E
Evaluate. Final answer: (2x + 3)(x + 4).
Check: (2x + 3)(x + 4) = 2x² + 8x + 3x + 12 = 2x² + 11x + 12 ✓
Lesson Summary
Monic quadratic — x² + bx + c
Find two numbers with product c and sum b. They become the constants in the brackets.
bcFactors
++both positive
−+both negative
+−mixed; positive larger
−−mixed; negative larger
Non-monic quadratic H — ax² + bx + c
AC method:
1
Compute ac
2
Find two numbers with product ac and sum b
3
Split the middle term
4
Factorise by grouping
Top 3 traps
Checking sum only, not product — both conditions must hold
Sign mismatch — brackets carry the exact signs of the numbers found
Skipping the check — always expand your answer to verify
Next lesson
Lesson 2.05 — Algebraic FractionsH: factorising quadratics becomes essential for simplifying fractions like (x² − 9)/(x² + 5x + 6).
Next: What You'll Learn