Factorisation
Introduction
Factorisation is the reverse process of expansion. While expansion 'removes brackets', factorisation 'introduces brackets' by expressing a polynomial as a product of its factors. This is a crucial skill for solving equations and simplifying expressions in ICSE Class 9.
Method 1: Common Factor
Take out the highest common factor (HCF) from all terms.
<ICSEExample title="Factorise 6x³ + 12x²"> <Solution> 6x³ + 12x² HCF of 6 and 12 is 6. HCF of x³ and x² is x². = 6x²(x + 2) </Solution> </ICSEExample> <ICSEExample title="Factorise 15a²b — 20ab²"> <Solution> 15a²b — 20ab² HCF of coefficients 15 and 20 is 5. HCF of a²b and ab² is ab. = 5ab(3a — 4b) </Solution> </ICSEExample>Method 2: Grouping
When an expression has four terms, group them into pairs, factor each pair, and take out the common bracket.
<ICSEExample title="Factorise x³ + 3x² + 2x + 6"> <Solution> x³ + 3x² + 2x + 6 = (x³ + 3x²) + (2x + 6) = x²(x + 3) + 2(x + 3) = (x + 3)(x² + 2) </Solution> </ICSEExample> <ICSEExample title="Factorise 2xy — 2xz + 3y — 3z"> <Solution> 2xy — 2xz + 3y — 3z = 2x(y — z) + 3(y — z) = (y — z)(2x + 3) </Solution> </ICSEExample>Method 3: Difference of Squares
Formula: a² — b² = (a + b)(a — b)
<ICSEExample title="Factorise 49x² — 36y²"> <Solution> 49x² — 36y² = (7x)² — (6y)² = (7x + 6y)(7x — 6y) </Solution> </ICSEExample> <ICSEExample title="Factorise 16a⁴ — 81b⁴"> <Solution> 16a⁴ — 81b⁴ = (4a²)² — (9b²)² = (4a² + 9b²)(4a² — 9b²) = (4a² + 9b²)(2a + 3b)(2a — 3b) </Solution> </ICSEExample>Method 4: Sum and Difference of Cubes
Sum of cubes: a³ + b³ = (a + b)(a² — ab + b²) Difference of cubes: a³ — b³ = (a — b)(a² + ab + b²)
<ICSEExample title="Factorise 8x³ + 27y³"> <Solution> 8x³ + 27y³ = (2x)³ + (3y)³ = (2x + 3y)[(2x)² — (2x)(3y) + (3y)²] = (2x + 3y)(4x² — 6xy + 9y²) </Solution> </ICSEExample> <ICSEExample title="Factorise 125 — 64x³"> <Solution> 125 — 64x³ = 5³ — (4x)³ = (5 — 4x)[5² + 5(4x) + (4x)²] = (5 — 4x)(25 + 20x + 16x²) </Solution> </ICSEExample>Method 5: Splitting the Middle Term
For a quadratic expression ax² + bx + c:
- Find two numbers p and q such that p + q = b and p × q = ac
- Split bx as px + qx
- Factorise by grouping
Method 6: Factor Theorem
Factor Theorem: If f(a) = 0 for a polynomial f(x), then (x — a) is a factor of f(x).
<ICSEExample title="Factorise x³ — 6x² + 11x — 6 using Factor Theorem"> <Solution> Let f(x) = x³ — 6x² + 11x — 6 Try x = 1: f(1) = 1 — 6 + 11 — 6 = 0. So (x — 1) is a factor. Divide f(x) by (x — 1): x³ — 6x² + 11x — 6 = (x — 1)(x² — 5x + 6) = (x — 1)(x — 2)(x — 3) </Solution> </ICSEExample>Common Mistakes With Fixes
| Mistake | Correction |
|---|---|
| Forgetting to take out common factor first | Always check for common factor before other methods |
| Wrong signs in difference of squares | a² — b² = (a+b)(a-b), not (a-b)(a-b) |
| Stopping too early in repeated factorisation | Keep factorising until all factors are prime |
| Incorrect sign when splitting middle term | Check that p × q = ac (including signs) |
ICSE Exam Focus
| Topic | Marks (approx.) | Frequency |
|---|---|---|
| Splitting middle term | 3-4 marks | Very common |
| Difference of squares | 3 marks | Common |
| Grouping | 3 marks | Common |
| Factor theorem | 4-5 marks | Frequently asked |
Self-Test
Q1: Factorise: 12x²y — 18xy²
Q2: Factorise: x² + 8x + 15
Q3: Factorise: 4x² — 25y²
Q4: Factorise: 3x² + 7x — 6
Q5: Factorise: x³ + 2x² — 5x — 6 (using Factor Theorem)
Q6: Factorise: 27a³ — 8b³
