Factorisation

1. What is Factorisation?

Factorisation is the process of writing an algebraic expression as a PRODUCT of its FACTORS.

6x² + 9x = 3x(2x + 3). The factors are 3x and (2x + 3).

'Factorisation is the REVERSE of expansion. While expansion REMOVES brackets, factorisation INTRODUCES them.'


2. Method 1 — Taking Common Factor

Find the HCF of ALL terms and factor it out.

Worked Example: Factorise 12x³y — 18x²y² + 6xy³

HCF of 12, 18, 6 is 6. HCF of x³, x², x is x. HCF of y, y², y³ is y. Common factor = 6xy.

12x³y — 18x²y² + 6xy³ = 6xy(2x² — 3xy + y²)

Worked Example: Factorise 4a²b — 6ab² + 8a²b²

Common factor = 2ab. = 2ab(2a — 3b + 4ab)


3. Method 2 — Factorisation by Grouping

When there is NO common factor in ALL terms, GROUP terms that have a common factor.

Steps:

  1. Group terms with common factors.
  2. Factor out the common factor from EACH group.
  3. Look for a COMMON BINOMIAL factor and factor it out.

Worked Example: Factorise 2xy + 3x + 2y + 3

Group: (2xy + 3x) + (2y + 3) Factor each group: x(2y + 3) + 1(2y + 3) Common binomial: (2y + 3) = (2y + 3)(x + 1)

Worked Example: Factorise a² + ab + ac + bc

Group: (a² + ab) + (ac + bc) = a(a + b) + c(a + b) = (a + b)(a + c)


4. Method 3 — Using Identities

Recognise expressions that match standard identities.

IdentityFactorised Form
a² + 2ab + b²(a + b)²
a² — 2ab + b²(a — b)²
a² — b²(a + b)(a — b)
x² + (a+b)x + ab(x + a)(x + b)

Worked Example: Factorise 16x⁴ — 81y⁴

= (4x²)² — (9y²)² = (4x² + 9y²)(4x² — 9y²) = (4x² + 9y²)(2x + 3y)(2x — 3y)


5. Method 4 — Splitting the Middle Term

For quadratic expressions of the form ax² + bx + c:

  1. Find two numbers p and q such that p + q = b AND p × q = ac
  2. Split the middle term bx as px + qx
  3. Factor by grouping

Worked Example: Factorise x² + 7x + 12

a = 1, b = 7, c = 12 Find p, q: p + q = 7, p × q = 12 p = 3, q = 4 (or 4, 3)

x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)

Worked Example: Factorise x² — 5x + 6

p + q = —5, p × q = 6 p = —2, q = —3

x² — 2x — 3x + 6 = x(x — 2) — 3(x — 2) = (x — 2)(x — 3)

Worked Example: Factorise 2x² + 7x + 3

a = 2, b = 7, c = 3 ac = 6. Find p, q: p + q = 7, p × q = 6 p = 6, q = 1

2x² + 6x + 1x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)

Worked Example: Factorise 6x² — x — 2

a = 6, b = —1, c = —2 ac = —12. Find p, q: p + q = —1, p × q = —12 p = —4, q = 3 (or p = 3, q = —4)

6x² — 4x + 3x — 2 = 2x(3x — 2) + 1(3x — 2) = (3x — 2)(2x + 1)


6. Choosing the Right Method

Expression TypeMethod
ALL terms share a factorCommon factor
Four or more terms with groupsGrouping
Perfect square trinomiala² ± 2ab + b² identity
Difference of two squaresa² — b² identity
Quadratic (x² + bx + c) where a = 1Split middle term
Quadratic (ax² + bx + c) where a ≠ 1Split middle term (find ac)
Four terms, two pairsGrouping

Common Mistakes and Fixes

MistakeFix
'Stopping too early — leaving a common factor'Check each factor: is it FULLY factorised?
'Sign errors in grouping'When factoring —(x — y), write as —x + y. Double-check signs
'Forgetting the 1 when factoring'When all terms cancel, keep 1: a + ab = a(1 + b), not a(b)
'Wrong splitting numbers'Always VERIFY: p + q = b AND p × q = ac. BOTH conditions must hold

ICSE Exam Focus (6–8 marks)

  • 2-mark questions: Factor out common factor
  • 3-mark questions: Factorise by grouping or using identities
  • 4-mark questions: Split the middle term (a = 1 or a ≠ 1)
  • 6-mark questions: Multi-method factorisation (three or more steps)

Self-Test

Q1. Factorise: 15ab — 10a²b² + 20ab² A1. Common factor = 5ab. = 5ab(3 — 2ab + 4b).

Q2. Factorise: 6xy — 4x + 9y — 6 A2. Group: (6xy — 4x) + (9y — 6) = 2x(3y — 2) + 3(3y — 2) = (3y — 2)(2x + 3).

Q3. Factorise: 9a² — 25b² A3. = (3a)² — (5b)² = (3a + 5b)(3a — 5b).

Q4. Factorise: x² + 9x + 20 A4. p + q = 9, p × q = 20. p = 4, q = 5. x² + 4x + 5x + 20 = x(x+4) + 5(x+4) = (x+4)(x+5).

Q5. Factorise: 3x² + 10x + 8 A5. ac = 24. p + q = 10, p × q = 24. p = 6, q = 4. 3x² + 6x + 4x + 8 = 3x(x+2) + 4(x+2) = (x+2)(3x+4).

Q6. Factorise: 4x² — 12x + 9 A6. (2x)² — 2(2x)(3) + 3² = (2x — 3)².

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