IGCSE Maths · Algebra and Graphs · Extended only

Advanced Factorising and Completing the Square Extended only

This entire note is Extended-only — Core stops at factorising by taking out a common factor. Everything here builds directly on that: you’re still looking for a common structure to pull out, just in cleverer disguises. These skills are essential later for solving quadratic equations and sketching curves, so it’s worth being genuinely fluent here rather than just recognising the patterns.

Why this matters Factorising a quadratic is how you solve equations like x² + 5x + 6 = 0 without a formula — and completing the square is how you find the exact turning point of a curve, or handle a quadratic that won’t factorise neatly at all. Both ideas come back constantly once equations and graphs begin.

1. Factorising by Grouping (Four-Term Expressions)

When an expression has four terms with no single common factor across all of them, look for a common factor in pairs instead.

ax + bx + kay + kby = x(a + b) + ky(a + b) = (a + b)(x + ky)
Worked example Factorise 5x + 5y + ax + ay fully.
Group in pairs: (5x + 5y) + (ax + ay)
Factorise each pair: 5(x + y) + a(x + y)
Both terms now share the bracket (x + y): (x + y)(5 + a)
Worked example — negative common factor in the second group Factorise 2pq + 2pr − 3q − 3r fully.
Group in pairs: (2pq + 2pr) + (−3q − 3r)
Factorise each pair: 2p(q + r) − 3(q + r)
Answer: (q + r)(2p − 3)
Try it — Factorising by Grouping
  1. Factorise 3x + 3y + bx + by fully.
  2. Factorise 4mn + 4mp − 3n − 3p fully.
  3. Factorise ax − ay + bx − by fully.
Show answers
  1. (x + y)(3 + b)
  2. (n + p)(4m − 3)
  3. (x − y)(a + b)

2. Difference of Two Squares

This is one of the most recognisable patterns in algebra — worth spotting instantly.

a²x² − b²y² = (ax − by)(ax + by)
Worked examples Factorise 9x² − 16y².
9x² = (3x)², 16y² = (4y)²
Answer: (3x − 4y)(3x + 4y)

Factorise x² − 49.
x² = (x)², 49 = (7)²
Answer: (x − 7)(x + 7)
Try it — Difference of Two Squares
  1. Factorise 25a² − 4b².
  2. Factorise x² − 64.
  3. Factorise 36p² − 81q².
  4. Factorise 100 − y².
Show answers
  1. (5a − 2b)(5a + 2b)
  2. (x − 8)(x + 8)
  3. (6p − 9q)(6p + 9q) — check further: 3 is a common factor of both terms first, so it’s cleaner as 9(4p²−9q²) = 9(2p−3q)(2p+3q); either fully-expanded form is valid, but the second is fully simplified.
  4. (10 − y)(10 + y)

3. Perfect Square Trinomials

A perfect square trinomial is what you get from squaring a bracket — recognise it by the first and last terms both being perfect squares, and the middle term being exactly double their square roots multiplied together.

a² + 2ab + b² = (a + b)²
a² − 2ab + b² = (a − b)²
Worked examples Factorise x² + 6x + 9.
Check: x² is a square, 9 = 3², and 6x = 2 × x × 3. ✓
Answer: (x + 3)²

Factorise 4x² − 12x + 9.
Check: 4x² = (2x)², 9 = 3², and 12x = 2 × 2x × 3. ✓
Answer: (2x − 3)²
Try it — Perfect Square Trinomials
  1. Factorise x² + 10x + 25.
  2. Factorise x² − 14x + 49.
  3. Factorise 9x² + 24x + 16.
Show answers
  1. (x + 5)²
  2. (x − 7)²
  3. (3x + 4)²

4. Factorising Quadratic Trinomials (ax² + bx + c)

When a = 1

Find two numbers that multiply to give c, and add to give b.

Worked example Factorise x² + 7x + 12.
Need two numbers that multiply to 12 and add to 7: 3 and 4
Answer: (x + 3)(x + 4)

When a ≠ 1 — the “split the middle term” method

Find two numbers that multiply to give a × c, and add to give b. Use them to split the middle term into two, then factorise by grouping.

Worked example Factorise 2x² + 7x + 3.
a × c = 2 × 3 = 6. Need two numbers multiplying to 6, adding to 7: 6 and 1
Rewrite: 2x² + 6x + x + 3
Group: 2x(x + 3) + 1(x + 3)
Answer: (x + 3)(2x + 1)
Worked example — with a negative constant Factorise 3x² − 5x − 2.
a × c = 3 × (−2) = −6. Need two numbers multiplying to −6, adding to −5: −6 and 1
Rewrite: 3x² − 6x + x − 2
Group: 3x(x − 2) + 1(x − 2)
Answer: (x − 2)(3x + 1)
Try it — Quadratic Trinomials
  1. Factorise x² + 9x + 20.
  2. Factorise x² − 3x − 10.
  3. Factorise 2x² + 11x + 5.
  4. Factorise 6x² − 7x − 3.
  5. Factorise 4x² − 4x − 15.
Show answers
  1. (x + 4)(x + 5)
  2. (x − 5)(x + 2)
  3. ac = 10, need numbers multiplying to 10 adding to 11: 10 and 1 → 2x²+10x+x+5 → 2x(x+5)+1(x+5) = (x + 5)(2x + 1)
  4. ac = −18, need numbers multiplying to −18 adding to −7: −9 and 2 → 6x²−9x+2x−3 → 3x(2x−3)+1(2x−3) = (2x − 3)(3x + 1)
  5. ac = −60, need numbers multiplying to −60 adding to −4: −10 and 6 → 4x²−10x+6x−15 → 2x(2x−5)+3(2x−5) = (2x − 5)(2x + 3)

5. Factorising Cubics (ax³ + bx² + cx)

Always take out the common factor first — this leaves a quadratic trinomial behind, which you can then factorise using the methods above.

Worked example Factorise x³ + 5x² + 6x fully.
Take out the common factor x: x(x² + 5x + 6)
Factorise the trinomial: x² + 5x + 6 = (x + 2)(x + 3)
Answer: x(x + 2)(x + 3)
Worked example Factorise 2x³ − 2x² − 12x fully.
Take out the common factor 2x: 2x(x² − x − 6)
Factorise the trinomial: x² − x − 6 = (x − 3)(x + 2)
Answer: 2x(x − 3)(x + 2)
Try it — Factorising Cubics
  1. Factorise x³ + 8x² + 15x fully.
  2. Factorise 3x³ − 6x² − 45x fully.
  3. Factorise x³ − 9x fully (hint: this is a common factor and a difference of two squares combined).
Show answers
  1. x(x + 3)(x + 5)
  2. 3x(x² − 2x − 15) = 3x(x − 5)(x + 3)
  3. x(x² − 9) = x(x − 3)(x + 3)

6. Completing the Square

Not every quadratic factorises neatly with whole numbers — completing the square rewrites any quadratic in the form (x + p)² + q, which is useful for solving equations and, later, for finding the exact turning point of a parabola.

When a = 1

x² + bx + c = (x + b/2)² − (b/2)² + c
Worked example Write x² + 6x + 5 in completed square form.
Half of 6 is 3. (x + 3)² expands to x² + 6x + 9 — that’s 4 more than we want, so subtract 4.
Answer: (x + 3)² − 4
Check: (x+3)² − 4 = x² + 6x + 9 − 4 = x² + 6x + 5 ✓

When a ≠ 1

Factor a out of the x² and x terms first, complete the square inside the bracket, then multiply back out carefully.

Worked example Write 2x² + 8x + 3 in completed square form.
Factor 2 out of the first two terms: 2(x² + 4x) + 3
Complete the square inside: x² + 4x = (x + 2)² − 4
Substitute back: 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 8 + 3
Answer: 2(x + 2)² − 5
Try it — Completing the Square
  1. Write x² + 8x + 10 in completed square form.
  2. Write x² − 4x + 1 in completed square form.
  3. Write 3x² + 12x + 7 in completed square form.
Show answers
  1. (x + 4)² − 6
  2. (x − 2)² − 3
  3. 3(x² + 4x) + 7 = 3[(x+2)² − 4] + 7 = 3(x+2)² − 12 + 7 = 3(x + 2)² − 5
Common mistakes to watch for
  • Grouping four-term expressions into the wrong pairs — try more than one pairing if the first doesn’t share a common bracket factor.
  • Spotting a “difference of two squares” pattern where the sign is actually a plus — a² + b² does not factorise this way (only a² − b² does).
  • For split-the-middle-term factorising, finding two numbers that add correctly but multiply incorrectly (or vice versa) — both conditions must hold.
  • Forgetting to take out the common factor first in a cubic — trying to factorise the whole cubic directly is much harder than factorising the quadratic left behind.
  • In completing the square with a ≠ 1, forgetting to multiply the “− (b/2)²” term back by a when substituting back in.

7. Quick-Fire Challenge Round

True or False?
  1. x² − 25 factorises as (x − 5)(x + 5).
  2. x² + 25 factorises using the difference of two squares.
  3. To factorise ax² + bx + c when a ≠ 1, the two numbers you need must multiply to give a × c.
  4. Completing the square only works when a = 1.
Show answers
  1. True.
  2. False — there’s no “sum of two squares” factorisation using real numbers; the difference of two squares only applies when it’s a subtraction.
  3. True.
  4. False — it works for any a, you just factor a out first and multiply back in at the end.
Spot the Error
  1. “x² + 8x + 16 factorises as (x + 4)(x − 4).”
  2. “To factorise 2x² + 5x + 3, I need two numbers that multiply to 5 and add to 6.”
  3. “Completing the square: x² + 10x + 3 = (x + 10)² + 3.”
Show answers
  1. x² + 8x + 16 is a perfect square trinomial (16 = 4², 8x = 2×4×x), so it factorises as (x + 4)², not (x + 4)(x − 4) — that expansion would give x² − 16, missing the middle term entirely.
  2. For 2x² + 5x + 3, the two numbers must multiply to a × c = 2 × 3 = 6, and add to b = 5 — the numbers are 2 and 3, not numbers multiplying to 5.
  3. Half of 10 is 5, not 10 — the correct completed square form is (x + 5)² − 25 + 3 = (x + 5)² − 22.

8. Mixed Practice — Bringing It All Together

  1. Factorise 4x + 4y + cx + cy fully.
  2. Factorise 16x² − 9y².
  3. Factorise x² + 12x + 36.
  4. Factorise x² − 2x − 24.
  5. Factorise 3x² + 14x − 5.
  6. Factorise x³ − 6x² + 8x fully.
  7. Write x² + 4x + 9 in completed square form.
Show answers
  1. (x + y)(4 + c)
  2. (4x − 3y)(4x + 3y)
  3. (x + 6)²
  4. (x − 6)(x + 4)
  5. ac = −15, numbers −1 and 15 → 3x²+15x−x−5 → 3x(x+5)−1(x+5) = (x + 5)(3x − 1)
  6. x(x² − 6x + 8) = x(x − 2)(x − 4)
  7. (x + 2)² + 5

Practice at Home

A large set again — take it steadily across a few sittings.

Part A — Factorising by grouping
  1. Factorise 7x + 7y + dx + dy fully.
  2. Factorise 5mn − 5mp + 2n − 2p fully.
  3. Factorise ax + 3a − bx − 3b fully.
Part B — Difference of two squares and perfect squares
  1. Factorise 49x² − 16y².
  2. Factorise x² − 121.
  3. Factorise x² + 20x + 100.
  4. Factorise 9x² − 30x + 25.
Part C — Factorising quadratic trinomials
  1. Factorise x² + 11x + 24.
  2. Factorise x² − 5x − 14.
  3. Factorise 2x² + 9x + 4.
  4. Factorise 5x² − 13x − 6.
Part D — Factorising cubics
  1. Factorise x³ + 7x² + 10x fully.
  2. Factorise 4x³ − 20x² + 24x fully.
  3. Factorise x³ − 16x fully.
Part E — Completing the square
  1. Write x² + 12x + 20 in completed square form.
  2. Write x² − 6x + 4 in completed square form.
  3. Write 2x² + 20x + 3 in completed square form.
Part F — Applied / challenge
  1. A square has side length (x + 5). Its area can be written as x² + 10x + 25. Confirm this by expanding (x + 5)², then explain what this shows about factorising perfect squares.
  2. Factorise 2x³ + 8x² − 10x fully — this one needs the common factor taken out before the trinomial can be split.
  3. Using completing the square, write x² + 5x + 2 in the form (x + p)² + q, where p and q are not whole numbers.

Key Vocabulary Recap

Factorise by grouping — pairing terms to find a shared bracket factor
Difference of two squares — a² − b² = (a − b)(a + b)
Perfect square trinomial — the result of squaring a two-term bracket
Trinomial — an expression with three terms, e.g. ax² + bx + c
Complete the square — rewrite a quadratic as (x + p)² + q

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section E2.2 (Extended subject content), points 4 and 5 — Factorising expressions of the form ax+bx+kay+kby, a²x²−b²y², a²+2ab+b², ax²+bx+c and ax³+bx²+cx, and completing the square. There is no Core equivalent — the corresponding Core section (C2.3) is explicitly marked “Extended content only” in the syllabus.

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