Algebra 4 of 6 Factorising
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🧩 Factorising — Algebra 4

Factorising is the reverse of expanding. Instead of removing brackets, you put them back in. It’s one of the most powerful tools in algebra — used to solve equations, simplify expressions, and reveal the structure hidden inside a formula.

Reverse of expanding Common factor (HCF) Quadratic factorising Difference of two squares Engineering applications
Factorising and expanding are exact opposites of each other 3(2x + 5) 6x + 15 EXPANDING → ← FACTORISING
What is factorising?Writing an expression as a product of its factors

Factorising means writing an expression as a multiplication — finding what multiplies together to give you the original expression.

6x + 15 = 3(2x + 5)
x² + 7x + 12 = (x + 3)(x + 4)
x² − 16 = (x + 4)(x − 4)
💡 Think of factorising like factoring a number: 12 = 3 × 4. You’re finding what multiplies together to make it. With algebra, you find what brackets multiply together.
Why do engineers factorise?
  1. Solve equations — x² + 7x + 12 = 0 is much easier to solve once factorised: (x+3)(x+4)=0
  2. Simplify fractions — cancel common factors top and bottom
  3. Spot structure — a²−b² = (a+b)(a−b) reveals it’s a difference of squares
  4. Engineering formulae — beam formulas often factorise neatly, making calculation faster
📌 In T Level papers, “factorise fully” means find the most simplified bracket form. Partially factorising loses marks.
Three types to masterEach has its own method
Type 1: Common factor — 6x² + 4x = 2x(3x + 2)
Look for the HCF of all terms and take it outside.
Type 2: Quadratic — x² + 7x + 12 = (x + 3)(x + 4)
Find two numbers that × to c and + to b.
Type 3: Difference of squares — x² − 25 = (x+5)(x−5)
Spot a²−b² pattern and apply the identity.
🧲

Common Factor Factorising

The simplest and most important type. Find the Highest Common Factor (HCF) of all terms, write it outside a bracket, then divide each term by it to find what goes inside.

Method: HCF outside the bracket × (remaining terms inside) = original expression
Factorise 12x² + 8x — step by step STEP 1 — Write the terms separately 12 x² + 8 x Coefficients: HCF(12, 8) = 4    Variables: both have x → HCF = 4x STEP 2 — Divide each term by the HCF (4x) 12x² ÷ 4x = 3x + 8x ÷ 4x = 2 Write HCF outside, remainders inside the bracket STEP 3 — Write the factorised answer 4x(3x + 2) ✅
Step-by-step method
  1. Find the HCF of the coefficients of all terms
  2. Find the highest power of each variable that appears in ALL terms
  3. Combine: HCF = (HCF of numbers) × (common variable part)
  4. Write HCF outside the bracket
  5. Divide each term by HCF — write results inside the bracket
  6. Check by expanding: should give the original expression
✅ Quick check: expand your answer — if you get back to the original, it’s correct!
Worked Examples
Factorise: 15x + 10
HCF: HCF(15,10) = 5, no variables
Divide: 15x÷5=3x   10÷5=2
Answer: 5(3x + 2) ✅
Factorise: 6x²y + 9xy² − 3xy
HCF nums: HCF(6,9,3) = 3
HCF vars: x appears in all (power 1), y appears in all (power 1)
HCF: 3xy
Divide: 6x²y÷3xy=2x   9xy²÷3xy=3y   3xy÷3xy=1
Answer: 3xy(2x + 3y − 1) ✅
⚠️ Factorise FULLY — always find the highest common factor, not just any common factor. 4x + 8 → 2(2x+4) is NOT fully factorised. The full answer is 4(x+2).
🧲 Interactive Tool — HCF Explorer TRY IT

Pick two coefficients using the dropdowns below. The tool will find the HCF step by step and show you the factorised form. Try different combinations!

📌 How to use: choose values from the dropdowns → watch the working appear live below → change values to explore more examples
🔢

Factorising Quadratics — x² + bx + c

When the expression has three terms and starts with x² (coefficient of 1), use the find two numbers method. The key is finding the pair that multiplies to give c and adds to give b.

Rule: x² + bx + c = (x + p)(x + q)   where   p × q = c   AND   p + q = b
Find two numbers that MULTIPLY to 12 and ADD to 7 for x² + 7x + 12 — c=12 (multiply to this), b=7 (add to this) Number pair Multiply = c(12)? Add = b(7)? Both conditions met? 1 and 12 1×12=12 ✔ 1+12=13 ❌ ❌ Add wrong — try again 2 and 6 2×6=12 ✔ 2+6=8 ❌ ❌ Add wrong — try again
3 and 4 3×4=12 ✔ 3+4=7 ✔ ✔✔ BOTH MATCH! → (x + 3)(x + 4)
Step-by-stepThe two-number method
  1. Identify b (coefficient of x) and c (constant term)
  2. List pairs of numbers that multiply to give c
  3. Find which pair also adds to give b
  4. Write the answer as (x + p)(x + q)
  5. Check by expanding (x+p)(x+q) — should give original
💡 Always start with the smallest factors. For c=20, try: 1×20, 2×10, 4×5. Stop when you find the pair that adds to b.
Worked Examples
Factorise: x² + 9x + 20
b=9, c=20
Pairs ×20: 1×20, 2×10, 4×5
4+5=9: ✔ found!
Answer: (x + 4)(x + 5) ✅
Factorise: x² − 5x + 6
b=−5, c=6 Both factors must be negative!
Pairs ×6: (−1)×(−6), (−2)×(−3)
−2+(−3): = −5 ✔
Answer: (x − 2)(x − 3) ✅
Factorise: x² + 2x − 15
b=2, c=−15 One pos, one neg!
Pairs ×−15: (−3)×5, (−3)×5 → −3+5=2 ✔
Answer: (x − 3)(x + 5) ✅
🔢 Interactive Tool — Quadratic Factoriser (x² + bx + c) TRY IT

Type values for b and c, then hit Factorise! — the tool searches for the number pair and shows every step of the working.

📌How to use: type b and c → click Factorise! → read through each step → change the numbers and try again
⚡

Harder Quadratics — ax² + bx + c (a ≠ 1)

When the coefficient of x² is not 1, use the AC method (also called the grouping method). It works every time!

AC Method: Find two numbers that × to a×c and + to b. Then split the middle term and factorise by grouping.
AC Method — step by step
  1. Identify a, b and c. Calculate a × c
  2. Find two numbers p, q where p × q = ac and p + q = b
  3. Rewrite the middle term bx as px + qx
  4. Factorise by grouping — group first two terms and last two terms
  5. Take out common factors from each group
  6. The bracket remaining should match — write the final answer
Worked: 2x² + 7x + 3
Factorise: 2x² + 7x + 3
a=2,b=7,c=3 ac = 2×3 = 6
Find p,q: p×q=6, p+q=7 → 1 and 6
Split 7x: 2x² + 1x + 6x + 3
Group: x(2x + 1) + 3(2x + 1)
Factor out: (2x + 1)(x + 3) ✅
Factorise: 3x² − 10x + 8
a=3,b=−10,c=8 ac = 24
p×q=24, p+q=−10 → −4 and −6
Split: 3x² − 4x − 6x + 8
Group: x(3x − 4) − 2(3x − 4)
Answer: (3x − 4)(x − 2) ✅
Factorise: 6x² + x − 2
a=6,b=1,c=−2 ac = −12
p×q=−12, p+q=1 → 4 and −3
Split: 6x² + 4x − 3x − 2
Group: 2x(3x + 2) − 1(3x + 2)
Answer: (3x + 2)(2x − 1) ✅
💡 Always check your answer by expanding! If (3x+2)(2x−1) = 6x²−3x+4x−2 = 6x²+x−2 ✔ — correct!
⬜

Difference of Two Squares

One of the most recognisable patterns in algebra. Whenever you see a squared term minus another squared term, it factorises instantly with no working needed.

a² − b² = (a + b)(a − b)    Always. No exceptions.
Difference of Two Squares — expand (a+b)(a−b) and watch the middle terms cancel THE PATTERN a² − b² = (a + b)(a − b) Expand (a+b)(a−b) using FOIL ↓ FIRST: a×a a² OUTER: a×(−b) −ab INNER: b×a +ab LAST: b×(−b) −b² −ab + ab = 0    The middle terms ALWAYS cancel! ❌❌ RESULT — only the squares survive a² − b² ✅
Spotting the patternTwo conditions must both be true
Condition 1: Both terms are perfect squares
Condition 2: They are subtracted (minus sign between them)
x² − 25 = x² − 5² → (x+5)(x−5) ✔
4x² − 9 = (2x)² − 3² → (2x+3)(2x−3) ✔
9a² − 16b² = (3a)² − (4b)² → (3a+4b)(3a−4b) ✔
⚠️ x² + 25 does NOT factorise — it’s a sum of squares, not a difference!
Worked Examples
Factorise: x² − 49
Spot it: x² − 7² (49 = 7²)
a=x, b=7
Answer: (x + 7)(x − 7) ✅
⚙ Engineering: Pipe area = πR² − πr²
Factorise: π(R² − r²)
DOTS: π(R+r)(R−r)
Answer: A = π(R+r)(R−r) ✅
⬜ Interactive Tool — Difference of Two Squares Builder TRY IT

Use the sliders to choose values for a and b. Watch the expression a²−b² build and factorise instantly — then see it verified by expanding back.

📌How to use: drag each slider left or right → the factorised form updates live → observe how a²−b² always gives two brackets with opposite signs
✔

Checking & Common Mistakes

One simple check catches all errors. And knowing the most common mistakes helps you avoid them before they cost you marks.

🔍
Spot the error — interactive checks Further down this page you'll find three expressions with answers. Decide for yourself whether each one is correct or wrong, then reveal the solution.
🤔Look at the expression and the given answer 🧮Try expanding it mentally or on paper 🔽Click Check it to see if you were right 📖Read the explanation carefully if you were wrong
The Golden CheckAlways expand your answer — it must give back the original
ORIGINAL EXPRESSION x² + 7x + 12 FACTORISED ANSWER (x + 3)(x + 4) FACTORISE → ← CHECK by expanding x²+4x+3x+12 = x²+7x+12 ✔ matches!
✅ Every time you factorise, expand your answer. If you get back to the original — correct! If not — find the error before moving on.
Common Mistakes
❌ Not factorising fully:
4x + 8 → 2(2x+4)   WRONG — 2x+4 can still be factorised!
✅ Correct: 4(x+2)
❌ Wrong signs in quadratics:
x²−5x+6 → (x+2)(x+3)   WRONG — signs must be negative!
✅ Correct: (x−2)(x−3)   (check: −2×−3=+6 ✔ −2+−3=−5 ✔)
❌ Confusing + and − in DOTS:
x²+25 → (x+5)(x+5)   WRONG — sum of squares doesn’t factorise!
✅ Only x²−25 factorises: (x+5)(x−5)
❌ Forgetting the −1 factor:
−x² + 4 → need to take out −1 first: −1(x²−4) = −1(x+2)(x−2)
Spot the ErrorsIs each one correct?

ℹ️ How to use: Work through each question and write down your answer. When you are ready, click Show Answer to reveal the full worked solution and mark scheme.

Check A
x² − 3x − 10 = (x − 5)(x + 2)
Expand: x²+2x−5x−10 = x²−3x−10 ✔
CORRECT ✅
Check B
x² + 6x + 9 = (x + 3)(x − 3)
Expand: x²−3x+3x−9 = x²−9 ≠ x²+6x+9 ❌
WRONG! Correct: (x+3)² = (x+3)(x+3)
Check C
6x² − 6 = 6(x² − 1) = 6(x+1)(x−1)
6(x+1)(x−1) = 6(x²−1) = 6x²−6 ✔
CORRECT ✅ (two-step: common factor then DOTS)
✅ Interactive Tool — Expand & Check Verifier TRY IT

Enter a factorised quadratic in the form (x + p)(x + q) using the number boxes. The tool expands it back and tells you whether it matches the original — the essential checking step!

📌How to use: enter p and q (the numbers inside your brackets) → click Check → see the full expansion and whether it confirms your factorisation
(x + )(x + )

💡 Tip: use negative numbers too! Try p = −5 and q = 2 to check (x−5)(x+2)

⚙️

Applications

Where this topic is used in engineering, manufacturing, maintenance and daily life.

⚙

Factorising in Engineering

Factorising appears directly in T Level / Level 3 Engineering when solving for unknown dimensions, simplifying stress formulas, and working with area and volume expressions.

🔨 Cross-Sectional Area

A hollow pipe: outer radius R, inner radius r. Wall area = πR² − πr². Factorise to get a useful form for calculations.

A = πR² − πr²
= π(R² − r²)
= π(R+r)(R−r)

R=50mm, r=44mm:
A = π(94)(6) ≈ 1772 mm²

⚡ Solving Power Equations

Power P = I²R − I²r = I²(R−r). Factorise to find current I when P and resistances are known.

P = I²R − I²r
= I²(R − r)

I² = P / (R−r)
I = √(P/(R−r))

📐 Finding Dimensions

A component has area x² + 9x + 20 mm². Factorise to find possible length and width expressions.

x² + 9x + 20 = (x+4)(x+5)

Length = (x+5) mm
Width = (x+4) mm
Check: (x+5)(x+4) = x²+9x+20 ✔

🏗️ Frame & Bracket Design

Volume of material in a frame: V = L²h − (L−2t)²h. Factorise to simplify calculation.

V = h[L² − (L−2t)²]
= h(L + L−2t)(L − L+2t)
= h(2L−2t)(2t)
= 4ht(L−t)
🔧 T Level Exam Style — Solve a quadratic in an engineering context
Problem: The cross-sectional area of a component is given by A = x² + 3x − 10. Find the value of x if A = 0. (This finds where the shape has zero area — a boundary condition.)
Factorise: x² + 3x − 10 = (x + 5)(x − 2)
Set = 0: (x + 5)(x − 2) = 0
Solve: x = −5 or x = 2
Context: x = 2 (discard −5 as a dimension cannot be negative) ✅
🎮

Quick Fire Quiz

Test your knowledge — 10 questions, instant feedback.

🎮
How to play You'll see a factorising question and four possible answers. Pick the one you think is correct — you get instant feedback after each choice.
👀Read the expression carefully 🖱️Click the answer you think is correct ✅Green = correct   🔴 Red = wrong ➡️Click Next to move to the next question 🔁Hit Play Again at the end to retry

⚡ Factorising Blitz

ℹ️ How to play: A question appears on screen. Click the answer you think is correct from the four options. Your score and streak update after each question. Click Next to move on.

Score: 0 Streak: 0 🔥 Q 1/10
Loading...
✏

Practice Questions

15 questions across 3 tiers. Always check your answer by expanding!

✏️
How to use these questions Work through each question on paper first — don't look at the answer until you've had a go. Then click the button to check.
📝Write your working on paper first 🔽Click Show answer to reveal the solution ✅Compare your method step by step 🔼Click Hide answer to close it again ⬆️Marks shown in blue — aim for full marks!
🟢 Tier 1 — Foundations
Q1
Factorise: 10x + 15
[2]
HCF(10,15)=5
10x÷5=2x   15÷5=3
= 5(2x + 3)
Q2
Factorise: 6x² − 9x
[2]
HCF = 3x
6x²÷3x=2x   9x÷3x=3
= 3x(2x − 3)
Q3
Factorise: x² + 8x + 15
[2]
Find two numbers that × to 15 and + to 8
Pairs ×15: 1×15, 3×5
3+5=8 ✔
= (x + 3)(x + 5)
Q4
Factorise: x² − 36
[2]
DOTS: x² − 6²
= (x + 6)(x − 6)
Q5
Factorise fully: 4x² − 8x
[2]
HCF = 4x
= 4x(x − 2)
🟡 Tier 2 — Core Skills
Q6
Factorise: x² − 7x + 12
[2]
b=−7, c=+12 → both negative
(−3)×(−4)=12   −3+−4=−7 ✔
= (x − 3)(x − 4)
Q7
Factorise: x² + x − 20
[2]
b=1, c=−20 → one pos, one neg
5×(−4)=−20   5+(−4)=1 ✔
= (x + 5)(x − 4)
Q8
Factorise: 3x² + 11x + 6
[3]
AC method: ac=18, b=11
2×9=18, 2+9=11 ✔
Split: 3x²+2x+9x+6
= x(3x+2)+3(3x+2)
= (3x + 2)(x + 3)
Q9
Factorise fully: 12x²y + 8xy − 4y
[3]
HCF = 4y
4y(3x² + 2x − 1)
Now factorise quadratic: ac=−3, b=2 → 3×(−1)
4y(3x²+3x−x−1)=4y[3x(x+1)−1(x+1)]
= 4y(3x − 1)(x + 1)
Q10
Factorise: 9x² − 25y²
[2]
DOTS: (3x)² − (5y)²
= (3x + 5y)(3x − 5y)
🔴 Tier 3 — T Level Challenge
Q11
Factorise: 6x² − 13x − 5
[3]
AC: ac=−30, b=−13
2×(−15)=−30, 2+(−15)=−13 ✔
6x²+2x−15x−5
= 2x(3x+1)−5(3x+1)
= (3x + 1)(2x − 5)
Q12
Factorise fully: 2x² − 50
[3]
Step 1: common factor → 2(x²−25)
Step 2: DOTS → 2(x+5)(x−5)
= 2(x + 5)(x − 5)
Q13
A rectangular component has area (2x² + 11x + 15) mm². Factorise to find expressions for the length and width.
[4]
AC: ac=30, b=11 → 5×6=30, 5+6=11 ✔
2x²+5x+6x+15
= x(2x+5)+3(2x+5)
= (2x+5)(x+3)
Length=(2x+5)mm, Width=(x+3)mm
Q14
Solve: x² − 4x − 12 = 0 by factorising.
[3]
b=−4, c=−12 → −6 and +2
(−6)×2=−12 ✔   −6+2=−4 ✔
(x−6)(x+2)=0
x = 6 or x = −2
Q15
A beam has cross-sectional area given by 4L² − 36. Factorise fully, then find L when the area = 0. Explain what this means in engineering context.
[5]
Step 1: 4(L²−9)
Step 2: 4(L+3)(L−3)
Set =0: L=3 or L=−3
Discard L=−3 (can’t have negative length)
L=3 means the cross-section has zero area at L=3 — this is the minimum viable beam size. ✅
🎯 Score guide: Q1–5 = foundations (10 marks) · Q6–10 = core skills (13 marks) · Q11–15 = T Level challenge (18 marks)

👉 Next topic: Solving Equations — using factorising and algebra to find exact values of unknowns.
SL

SkillLondon — T Level / Level 3 Engineering Maths  ·  Algebra 4: Factorising  ·  Next: Solving Equations →