Algebraic Identities
1. What is an Identity?
An identity is an equation that is TRUE for ALL values of the variable.
(a + b)² = a² + 2ab + b² is an identity — it holds for ANY a and b.
'An identity is DIFFERENT from an equation. An equation is true for only SPECIFIC values. An identity is true for ALL values.'
2. Standard Identities
Identity 1: (a + b)² = a² + 2ab + b²
The square of a SUM = square of first + twice the product + square of second.
Worked Example: Expand (3x + 4y)²
(3x)² + 2(3x)(4y) + (4y)² = 9x² + 24xy + 16y²
Worked Example: Find (x + 5)² using the identity.
x² + 2(x)(5) + 5² = x² + 10x + 25
Identity 2: (a — b)² = a² — 2ab + b²
The square of a DIFFERENCE = square of first — twice the product + square of second.
Worked Example: Expand (2x — 3y)²
(2x)² — 2(2x)(3y) + (3y)² = 4x² — 12xy + 9y²
Worked Example: Evaluate (97)² using identity.
97 = 100 — 3 (100 — 3)² = 10000 — 600 + 9 = 9409
Identity 3: a² — b² = (a + b)(a — b)
Difference of squares = product of SUM and DIFFERENCE.
Worked Example: Factorise 4x² — 9y²
4x² — 9y² = (2x)² — (3y)² = (2x + 3y)(2x — 3y)
Worked Example: Evaluate (102 × 98) using identity.
102 × 98 = (100 + 2)(100 — 2) = 100² — 2² = 10000 — 4 = 9996
Identity 4: (x + a)(x + b) = x² + (a + b)x + ab
Worked Example: Expand (x + 5)(x + 3)
x² + (5 + 3)x + 5×3 = x² + 8x + 15
Worked Example: Find (y — 3)(y + 7)
y² + (—3 + 7)y + (—3)(7) = y² + 4y — 21
3. Applications of Identities
Evaluating Squares and Products
Worked Example: Find 105² using identity.
105 = 100 + 5 105² = (100 + 5)² = 10000 + 2(100)(5) + 25 = 10000 + 1000 + 25 = 11025
Worked Example: Find 199² — 198²
199² — 198² = (199 + 198)(199 — 198) = 397 × 1 = 397
4. Factorisation Using Identities
Factorising a² + 2ab + b²
Recognise as (a + b)².
Worked Example: Factorise 4x² + 12x + 9
4x² = (2x)², 9 = 3², 2(2x)(3) = 12x ✓ 4x² + 12x + 9 = (2x + 3)²
Factorising a² — 2ab + b²
Recognise as (a — b)².
Worked Example: Factorise 25a² — 30ab + 9b²
25a² = (5a)², 9b² = (3b)², 2(5a)(3b) = 30ab ✓ 25a² — 30ab + 9b² = (5a — 3b)²
Factorising a² — b²
Use (a + b)(a — b).
Worked Example: Factorise 49x² — 16y²
49x² — 16y² = (7x + 4y)(7x — 4y)
5. Extension — (a + b + c)²
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Worked Example: Expand (2x + 3y — z)²
= (2x)² + (3y)² + (—z)² + 2(2x)(3y) + 2(3y)(—z) + 2(—z)(2x) = 4x² + 9y² + z² + 12xy — 6yz — 4zx
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| '(a + b)² = a² + b²' | (a + b)² = a² + 2ab + b². The MIDDLE term is CRUCIAL |
| '(—a — b)² = —(a + b)²' | (—a — b)² = [—(a+b)]² = (a + b)² = a² + 2ab + b² |
| 'a² — b² = (a — b)²' | (a — b)² = a² — 2ab + b². a² — b² = (a+b)(a—b) — NO middle term |
| 'Confusing (x+a)(x+b) with (x+a)(x—a)' | (x+a)(x—a) = x² — a² (Identity 3). (x+a)(x+b) = x² + (a+b)x + ab (Identity 4) |
ICSE Exam Focus (5–7 marks)
- 2-mark questions: Expand using identities
- 3-mark questions: Evaluate products/squares using identities
- 4-mark questions: Factorise expressions using identities
- 6-mark questions: Combined problems — simplify then evaluate
Self-Test
Q1. Expand (5a — 3b)². A1. 25a² — 30ab + 9b².
Q2. Factorise 9x² — 24xy + 16y². A2. (3x)² — 2(3x)(4y) + (4y)² = (3x — 4y)².
Q3. Find 47² using identity. A3. 47 = 50 — 3. (50—3)² = 2500 — 300 + 9 = 2209.
Q4. Factorise 36x² — 121y². A4. (6x)² — (11y)² = (6x + 11y)(6x — 11y).
Q5. Simplify: (x + 2)(x + 5) — (x — 2)(x — 5). A5. (x+2)(x+5) = x² + 7x + 10. (x—2)(x—5) = x² — 7x + 10. Difference = 14x.
Q6. Find the value of (203 × 197) using identity. A6. 203 × 197 = (200 + 3)(200 — 3) = 200² — 3² = 40000 — 9 = 39991.
