Solving Quadratic Equations by Factorising
SPEC A18 • ALGEBRA
⭐ High-priority. Foundation (Additional) + Higher. Master the zero-product property, spot hidden quadratics, handle rearrangement, and read approximate solutions from a graph.
Why This Topic Matters
Every GCSE paper includes at least one quadratic to solve. Factorising is the primary method for "nice" quadratics with integer (or simple fractional) roots. Completing the square and the quadratic formula follow in Lesson 2.22 — but factorising is almost always the fastest route when available.
Authentic exam evidence
Jun 2024 P3H Q19: "Solve (x − 3)² = 4x − 7" — 4 marks, requires rearrangement before factorising.
Nov 2023 P3H Q20: "Solve 2x(x + 10) = 5x − 18" — 4 marks, expand + rearrange + factorise non-monic.
Jun 2024 P3H Q13: "Write down the roots of the equation f(x) = 0" — 2 marks, reading approximate roots from a quadratic graph.
Tier note
Foundation (Additional): solve simple quadratics by factorising; find approximate roots from a graph. Higher only: solve quadratics that require rearrangement first.
Key Terms
Quadratic equation
An equation of the form ax ² + bx + c = 0, where a ≠ 0. The highest power of the unknown is 2.
Root / solution
A value of x that makes the equation true. A quadratic has up to two real roots (sometimes one repeated, sometimes none, but at GCSE you'll almost always have two distinct roots).
Zero-product property
If A × B = 0, then A = 0 or B = 0 (or both). This is the entire reason factorising works — a product is zero only if one of the factors is zero.
The Zero-Product Property
This is the idea that makes the whole method work. If a product equals zero, at least one of the things being multiplied must itself be zero.
Why it's so important
Without the zero-product property, factorising would be pointless. It's the bridge that converts "a product equals zero" into "solve two simple linear equations."
Only works when = 0
The property only works if the product equals zero. If (x + 5)(x − 2) = 8, you cannot set each factor to 8. You must first rearrange to make one side = 0, then factorise.
Standard 4-Step Procedure
1Rearrange so the equation is in the form ax² + bx + c = 0 (everything on one side, zero on the other).
2Factorise the quadratic expression. Skills from Lesson 2.04 apply — monic, non-monic, DOTS, common factor.
3Set each factor equal to zero. This produces two simple linear equations.
4Solve each linear equation for x. Report both solutions.
Monic Example: x ² + 3 x − 10 = 0
A "monic" quadratic has coefficient 1 on x². Factorising is fastest: find two numbers with product c and sum b.
Check both roots
Always verify: (−5)² + 3(−5) − 10 = 25 − 15 − 10 = 0 ✓. And (2)² + 3(2) − 10 = 4 + 6 − 10 = 0 ✓. Both roots confirmed.
Difference of Two Squares (DOTS)
Any equation of the form x² − k² = 0 factorises using the DOTS pattern: a ² − b ² = ( a + b )( a − b ).
Don't lose the negative root
Common mistake: writing "x² = 9, so x = 3" and missing x = −3. A quadratic has up to two real roots — (−3)² = 9 as well as 3² = 9, so both satisfy the equation. Using the factorising route reminds you to check both factors.
DOTS in disguise
Expressions like 4x² − 25 = 0 are DOTS too: (2x)² − 5² = (2x + 5)(2x − 5) = 0, giving x = ±5/2.
Non-Monic: 2 x ² + 7 x + 3 = 0
When the x² coefficient is not 1, you need the ac-method or "guess and check". For 2x² + 7x + 3, one bracket starts with 2x, the other with x.
ac-method alternative
Multiply a × c = 2 × 3 = 6. Find two numbers with product 6 and sum 7: that's 6 and 1. Split the middle: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Same answer.
Rearrange Before You Factorise Higher
Many Higher questions give you a quadratic that isn't yet in standard form. You must first expand and rearrange to get ax² + bx + c = 0 before factorising.
Example types
Given formRearrange to standard form(x − 3)² = 4x − 7x² − 10x + 16 = 02x(x + 10) = 5x − 182x² + 15x + 18 = 0x² = 5xx² − 5x = 0x(x − 4) = 5x² − 4x − 5 = 0
Two marks often go before you factorise
On 4-mark questions like Jun 2024 P3H Q19, the first two marks are earned before you factorise: M1 for correct expansion, M1dep for correct rearrangement to a three-term quadratic. Getting factorising right on a mis-rearranged equation earns only the third method mark.
Approximate Solutions from a Graph
The spec explicitly requires you to find approximate solutions using a graph. The x-intercepts of the parabola y = ax² + bx + c are the roots of ax² + bx + c = 0.
The parabola y = x² + x − 2 crosses the x-axis at x = −2 and x = 1 — these are the roots of the equation x² + x − 2 = 0.
Why "approximate"?
Unless the roots land on integer gridlines, you can only read them to the accuracy of the grid (typically to 1 decimal place). Algebra gives exact answers; graphs give visual confirmation or an estimate when algebra is hard.
MCE Chain for Quadratics
Every factorising question splits marks across these three stages. Know where the marks live.
MMethod. Identify the quadratic's form (monic, non-monic, DOTS, or hidden). If it's not already = 0, expand and rearrange first. State the factorisation strategy.
CCalculate. Factorise correctly. Set each factor to zero. Solve both linear equations.
EEvaluate. Report both roots (they're both required — missing one loses the A mark). Substitute back to verify where possible.
Worked Example: Setup
Ex
Higher — 4 marks (Jun 2024 P3H Q19 authentic)
Solve (x − 3)² = 4x − 7.
M — Method
The equation isn't in standard form yet. Two rearrangement steps needed:
Expand the left-hand side: (x − 3)² = x² − 6x + 9.
Rearrange everything to one side = 0.
Factorise the resulting quadratic.
Set each factor = 0 and solve.
Worked Example: Calculate & Evaluate
C — Calculate
(x − 3)² = 4x − 7
x² − 6x + 9 = 4x − 7 (expand LHS)
x² − 10x + 16 = 0 (subtract 4x, add 7)
Factorise: product 16, sum −10 → −2 and −8
(x − 2)(x − 8) = 0
x − 2 = 0 → x = 2
x − 8 = 0 → x = 8
E — Evaluate (check)
At x = 2: LHS = (2 − 3)² = 1. RHS = 4(2) − 7 = 1 ✓
At x = 8: LHS = (8 − 3)² = 25. RHS = 4(8) − 7 = 25 ✓
x = 2 and x = 8
Mark breakdown (Jun 2024 mark scheme)
M1 for (x−3)² = 4x−7 or x² − 6x + 9 = 4x − 7 (expansion). M1dep for x² − 10x + 16 (= 0) (rearrangement with terms collected). M1 for (x − 2)(x − 8) (factorisation). A1 for x = 2 and x = 8.
Weak vs Strong: Dividing by x
The single most common trap in factorising quadratics is dividing both sides by x when x could be zero. Happens most often when the equation looks like ax² = bx.
Weak student
Solves 2x² = 5x by dividing by x:
Missed x = 0.
Strong student
Rearranges and factorises:
Both roots recovered.
Mnemonic
See x on both sides? Subtract — don't divide. Dividing only works if you're certain the divisor isn't zero. With x, you're never certain.
Examiner Alerts
Only one root given
A quadratic typically has two solutions. Solving only one factor and stopping loses the A1 mark. Always set both factors = 0.
Dividing by x
On x² = 5x (or any equation where x is a factor both sides), never divide. You'd lose x = 0. Rearrange: x² − 5x = 0 → x(x − 5) = 0.
Ignoring the negative root
For x² = 9, students often write x = 3 and stop. But (−3)² = 9 too. Always check both factors when factorising DOTS: (x+3)(x−3) = 0.
Applying zero-product to non-zero RHS
If (x + 5)(x − 2) = 8, you cannot just set each factor equal to 8. Rearrange first: x² + 3x − 10 = 8 → x² + 3x − 18 = 0, then factorise the new quadratic.
Practice Question 1
Monic quadratic. Solve x² + 5x + 6 = 0. [2 marks]
M
Find two numbers with product 6 and sum 5: +2 and +3.
C
Factorise: (x + 2)(x + 3) = 0. Set each factor = 0: x + 2 = 0 → x = −2. x + 3 = 0 → x = −3.
E
x = −2 or x = −3. Check: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0 ✓; (−3)² + 5(−3) + 6 = 9 − 15 + 6 = 0 ✓. Marks: M1 factorisation, A1 both roots.
Practice Question 2
Hidden quadratic (the trap). Solve x² = 4x. [2 marks]
M
Rearrange to make one side = 0. Do NOT divide by x — you'd lose x = 0.
C
x² − 4x = 0 → factor out x: x(x − 4) = 0. Set each factor = 0: x = 0 or x − 4 = 0 → x = 4.
E
x = 0 or x = 4. Check: 0² = 0 and 4(0) = 0 ✓; 4² = 16 and 4(4) = 16 ✓.
Practice Question 3 Higher
Rearrangement + non-monic (Nov 2023 P3H Q20 authentic). Solve 2x(x + 10) = 5x − 18. [4 marks]
M
Expand LHS, then move everything to one side to get a standard-form quadratic.
C
2x(x + 10) = 5x − 18 → 2x² + 20x = 5x − 18 → 2x² + 15x + 18 = 0. Factorise: (2x + 3)(x + 6) = 0. Set each = 0: 2x + 3 = 0 → x = −3/2; x + 6 = 0 → x = −6.
E
x = −1.5 or x = −6. Mark structure: M1 for expansion (2x² + 20x); M1dep for rearrangement to 2x² + 15x + 18 = 0; M1 for factorisation (or quadratic formula); A1 for both roots.
Summary
The Zero-Product Property
If A × B = 0, then A = 0 or B = 0.
This is why factorising works — only valid when the equation equals 0.
4-Step Procedure
1
Rearrange to ax ² + bx + c = 0.
2
Factorise the quadratic (monic / non-monic / DOTS / common factor).
3
Set each factor = 0.
4
Solve both linear equations — report BOTH roots.
Foundation (Additional)
Monic quadratics
DOTS cases
Approximate roots from a graph
Higher extensions
Quadratics requiring rearrangement
Non-monic factorising
Hidden quadratics with the x = 0 trap
Top Exam Tip
Always rearrange to = 0 before factorising. Never divide by x. Never stop after one root — quadratics have up to two. And always check by substituting both roots back into the original equation.
Next: Why This Topic Matters