Exercise 12.2 — Factorisation by Identities

Factorisation using algebraic identities.

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Factorisation Using Algebraic Identities — The Four Key Formulas

Exercise 12.2 introduces a powerful upgrade to factorisation: instead of always hunting for a common factor, you can recognise the pattern of an expression and match it to one of four algebraic identities. Each identity is a reverse expansion formula — the right-hand side is the expanded form, and the left-hand side is the factorised form you want.

These identities are among the most important algebraic tools in Class 8, 9, and 10 mathematics and appear throughout CBSE, Telangana, and Andhra Pradesh board examinations.

Identity 1 — Perfect Square (Plus)

a² + 2ab + b² = (a + b)²
= (a + b)(a + b)

Identity 2 — Perfect Square (Minus)

a² − 2ab + b² = (a − b)²
= (a − b)(a − b)

Identity 3 — Difference of Squares

a² − b² = (a + b)(a − b)

Identity 4 — Trinomial Form

x² + (a+b)x + ab
= (x + a)(x + b)

💡 The pattern-matching principle: When you see a quadratic expression, ask: "Does it look like one of these four patterns?" If yes, read off the values of a, b (and x), then write the factorised form directly. No lengthy steps needed once you spot the pattern.

Question 1 — Factorise Using Perfect Square Identities

In Q1, every expression is either a perfect square trinomial (Identity 1 or 2) or can be reduced to one. The three-step method is: identify the squares of the first and last terms, check whether the middle term equals 2 × first × last, then write the square of a binomial.

🔍 How to spot a perfect square trinomial

Is the first term a perfect square? Write it as a².
Is the last term a perfect square? Write it as b².
Is the middle term exactly 2ab (positive → Identity 1) or −2ab (negative → Identity 2)?
If all three: write (a + b)² or (a − b)² directly.

Q1 (i) Identity 1: a² + 2ab + b²

Factorise: a² + 10a + 25

Identify the parts

a² → a = a
25 = 5² → b = 5
2ab = 2×a×5 = 10a ✓

Apply Identity 1

a² + 2(a)(5) + 5²
= (a + 5)²

∴ a² + 10a + 25 = (a + 5)(a + 5)

Q1 (ii) Identity 2: a² − 2ab + b²

Factorise: l² − 16l + 64

Identify the parts

l² → a = l
64 = 8² → b = 8
−2ab = −2×l×8 = −16l ✓

Apply Identity 2

l² − 2(l)(8) + 8²
= (l − 8)²

∴ l² − 16l + 64 = (l − 8)(l − 8)

Q1 (iii) Identity 1: a² + 2ab + b²

Factorise: 36x² + 96xy + 64y²

Identify the parts

36x² = (6x)² → a = 6x
64y² = (8y)² → b = 8y
2ab = 2×6x×8y = 96xy ✓

Apply Identity 1

(6x)² + 2(6x)(8y) + (8y)²
= (6x + 8y)²

∴ 36x² + 96xy + 64y² = (6x + 8y)(6x + 8y)

📌 The key skill here is recognising that 36x² = (6x)² and 64y² = (8y)². Always check if the coefficients of x² and y² are perfect squares before applying the identity.

Q1 (iv) Identity 2: a² − 2ab + b²

Factorise: 25x² + 9y² − 30xy

Rewrite in standard order: 25x² − 30xy + 9y²

25x² = (5x)² → a = 5x 9y² = (3y)² → b = 3y −2ab = −2×5x×3y = −30xy ✓

∴ 25x² − 30xy + 9y² = (5x − 3y)(5x − 3y)

⛔ Reordering alert: The expression is written as 25x² + 9y² − 30xy, with terms out of standard order. Always rearrange to a² − 2ab + b² order before matching the identity. Failing to reorder is a common exam mistake.

Q1 (v) Identity 2: a² − 2ab + b²

Factorise: 25m² − 40mn + 16n²

Identify the parts

25m² = (5m)² → a = 5m
16n² = (4n)² → b = 4n
−2ab = −2×5m×4n = −40mn ✓

Apply Identity 2

(5m)² − 2(5m)(4n) + (4n)²
= (5m − 4n)²

∴ 25m² − 40mn + 16n² = (5m − 4n)(5m − 4n)

Q1 (vi) Identity 2: a² − 2ab + b²

Factorise: 81x² − 198xy + 121y²

Identify the parts

81x² = (9x)² → a = 9x
121y² = (11y)² → b = 11y
−2ab = −2×9x×11y = −198xy ✓

Apply Identity 2

(9x)² − 2(9x)(11y) + (11y)²
= (9x − 11y)²

∴ 81x² − 198xy + 121y² = (9x − 11y)(9x − 11y)

Q1 (vii) Identity 2 (after expansion)

Factorise: (x + y)² − 4xy

1 Expand (x + y)²: = x² + 2xy + y² − 4xy

2 Combine like terms (2xy − 4xy = −2xy): = x² − 2xy + y²

3 Apply Identity 2 with a = x, b = y: = (x − y)(x − y)

💡 This question cleverly disguises a perfect square! First expand, then simplify the middle terms to reveal the identity. You cannot directly apply the identity to (x + y)² − 4xy without expanding first.

Q1 (viii) Identity 1 (higher powers)

Factorise: l⁴ + 4l²m² + 4m⁴

Identify the parts

l⁴ = (l²)² → a = l²
4m⁴ = (2m²)² → b = 2m²
2ab = 2×l²×2m² = 4l²m² ✓

Apply Identity 1

(l²)² + 2(l²)(2m²) + (2m²)²
= (l² + 2m²)²

∴ l⁴ + 4l²m² + 4m⁴ = (l² + 2m²)(l² + 2m²)

📌 When the expression has degree 4 variables (l⁴, m⁴), treat l² as a single unit 'a' and 2m² as 'b'. The identity works for any algebraic expressions, not just simple variables.

Question 2 — Factorise Using Difference of Squares (a² − b²)

Identity 3 — the Difference of Two Squares — applies to any expression of the form a² − b², giving the factorised form (a + b)(a − b). This is one of the most frequently tested identities in board examinations. The key skill is writing both terms as perfect squares, then reading off a and b.


  a² − b² = (a + b)(a − b)


  Rule: always check if BOTH terms are perfect squares with a MINUS sign between them.

Q2 (i) Identity 3: a² − b²

Factorise: x² − 36

x² − 36 = x² − 6² ← 36 = 6² a = x, b = 6 = (x + 6)(x − 6)

Q2 (ii) Identity 3: a² − b²

Factorise: 49x² − 25y²

49x² − 25y² = (7x)² − (5y)² ← 49 = 7², 25 = 5² a = 7x, b = 5y = (7x + 5y)(7x − 5y)

Q2 (iii) Identity 3: a² − b²

Factorise: m² − 121

m² − 121 = m² − 11² ← 121 = 11² a = m, b = 11 = (m + 11)(m − 11)

Q2 (iv) Identity 3: a² − b²

Factorise: 81 − 64x²

81 − 64x² = 9² − (8x)² ← 81 = 9², 64 = 8² a = 9, b = 8x = (9 + 8x)(9 − 8x)

📌 The constant term (81) comes first here. The identity a² − b² still applies — just let a = 9 (the number) and b = 8x (the variable term).

Q2 (v) Identity 3: a² − b²

Factorise: x²y² − 64

x²y² − 64 = (xy)² − 8² ← x²y² = (xy)², 64 = 8² a = xy, b = 8 = (xy + 8)(xy − 8)

Q2 (vi) Identity 3 (with common factor first)

Factorise: 6x² − 54

1 Take out common factor 6: = 6(x² − 9)

2 Apply Identity 3 to (x² − 9) = x² − 3²: a = x, b = 3 = 6(x + 3)(x − 3)

💡 Two-stage factorisation: First remove the common factor (6), then apply the identity to the remaining bracket. This is a very common question type in board exams — never skip the common-factor step.

Q2 (vii) Identity 3: a² − b²

Factorise: x² − 81

x² − 81 = x² − 9² ← 81 = 9² a = x, b = 9 = (x + 9)(x − 9)

Q2 (viii) Identity 3 applied twice

Factorise: 2x − 32x⁵

1 Take out common factor 2x: = 2x(1 − 16x⁴)

2 Write 1 = 1² and 16x⁴ = (4x²)²: = 2x[1² − (4x²)²]

3 Apply Identity 3 with a = 1, b = 4x²: = 2x(1 + 4x²)(1 − 4x²)

4 Apply Identity 3 again to (1 − 4x²) = 1² − (2x)²: = 2x(1 + 4x²)(1 + 2x)(1 − 2x)

💡 Three levels of factorisation! This is the most complex question in Q2. Each time you see a difference of squares, apply the identity again. Only stop when neither factor can be factorised further.

Q2 (ix) Common factor + Identity 3

Factorise: 81x⁴ − 121x²

1 Take out common factor x²: = x²(81x² − 121)

2 Write 81x² = (9x)² and 121 = 11²: = x²[(9x)² − 11²]

3 Apply Identity 3 with a = 9x, b = 11: = x²(9x + 11)(9x − 11)

Q2 (x) Identity 2 + Identity 3 combined

Factorise: p² − 2pq + q² − r²

1 Group first three terms — recognise Identity 2: p² − 2pq + q² = (p − q)²

2 Expression becomes (p − q)² − r²

3 Apply Identity 3 with a = (p − q), b = r: = (p − q + r)(p − q − r)

✅ This is a two-identity chain: first use Identity 2 to compress three terms into a square, then use Identity 3 on the resulting difference of squares. This combination appears frequently in higher-level questions.

Q2 (xi) Identity 3 with compound expressions

Factorise: (x + y)² − (x − y)²

a = (x + y), b = (x − y) = [(x+y) + (x−y)] × [(x+y) − (x−y)]

= [x + y + x − y] × [x + y − x + y] = [2x] × [2y] = 4xy

📌 When a and b are themselves compound expressions like (x + y), substitute them directly into the (a + b)(a − b) template, then simplify. The result here is simply 4xy — a surprisingly clean answer from a complex-looking question.

Question 3 — Mixed Factorisation (Common Factor and Grouping)

Question 3 combines techniques: some parts need a simple common factor, others need regrouping, and some need both. These are shorter problems designed to sharpen your ability to choose the right method quickly.

Q3 (i)

Factorise: lx² + mx

HCF = x = x(lx) + x(m) = x(lx + m)

Q3 (ii)

Factorise: 7y² + 35z²

HCF = 7 = 7(y²) + 7(5z²) = 7(y² + 5z²)

📌 Note: (y² + 5z²) cannot be factorised further using the identities above — it is a sum, not a difference of squares. The factorisation stops here.

Q3 (iii)

Factorise: 3x⁴ + 6x³y + 9x²z

HCF = 3x² = 3x²(x²) + 3x²(2xy) + 3x²(3z) = 3x²(x² + 2xy + 3z)

Q3 (iv)

Factorise: x² − ax − bx + ab

1 Group: (x² − ax) + (−bx + ab) 2 Factor each pair: = x(x − a) + b(−x + a) = x(x − a) − b(x − a) = (x − a)(x − b)

Q3 (v)

Factorise: 3ax − 6ay − 8by + 4bx

1 Group: (3ax − 6ay) + (4bx − 8by) 2 Factor each pair: = 3a(x − 2y) + 4b(x − 2y) = (3a + 4b)(x − 2y)

Q3 (vi)

Factorise: mn + m + n + 1

1 Group: (mn + m) + (n + 1) 2 Factor each pair: = m(n + 1) + 1(n + 1) = (m + 1)(n + 1)

💡 When a group has no variable to factor out, factor out 1. Here the second group (n + 1) has 1 as its factor, making the bracket (n + 1) visible in both groups.

Q3 (vii)

Factorise: 6ab − b² + 12ac − 2bc

1 Group: (6ab − b²) + (12ac − 2bc) 2 Factor each pair: = b(6a − b) + 2c(6a − b) = (6a − b)(b + 2c)

Q3 (viii)

Factorise: p²q − pr² − pq + r²

1 Group: (p²q − pr²) + (−pq + r²) 2 Factor each pair: = p(pq − r²) − 1(pq − r²) = (p − 1)(pq − r²)

Q3 (ix)

Factorise: x(y + z) − 5(y + z)

Common bracket = (y + z) = (y + z)(x) − (y + z)(5) = (y + z)(x − 5)

✅ This is the simplest form of bracket factorisation — the common factor is already a bracket (y + z). Simply take it out and write the remaining factors inside a new bracket.

Question 4 — Advanced Factorisation (Chaining Identities)

Question 4 problems require applying identities more than once, or combining two different identities in sequence. These are the most challenging questions in Exercise 12.2 and frequently appear as 3-mark or 4-mark questions in TS/AP and CBSE board papers.

Q4 (i) Identity 3 → Identity 3

Factorise: x⁴ − y⁴

1 Write x⁴ = (x²)² and y⁴ = (y²)²: = (x²)² − (y²)²

2 Apply Identity 3 with a = x², b = y²: = (x² + y²)(x² − y²)

3 Apply Identity 3 again to (x² − y²): = (x² + y²)(x + y)(x − y)

📌 (x² + y²) cannot be factorised further (sum of squares has no real factors). Only the difference (x² − y²) can be factorised again.

Q4 (ii) Identity 3 → Identity 1

Factorise: a⁴ − (b + c)⁴

1 Write as (a²)² − [(b+c)²]²: = [a² + (b+c)²][a² − (b+c)²]

2 Apply Identity 3 to second bracket (a = a, b = b+c): = [a² + (b+c)²][a + (b+c)][a − (b+c)]

3 Expand (b+c)² in first bracket: First bracket: a² + b² + c² + 2bc

= (a² + b² + c² + 2bc)(a + b + c)(a − b − c)

Q4 (iii) Identity 3 with compound b

Factorise: l² − (m − n)²

a = l, b = (m − n) = [l + (m − n)][l − (m − n)] = [l + m − n][l − m + n] = (l + m − n)(l − m + n)

Q4 (iv) Identity 3 with fraction

Factorise: 49x² − 16/25

49x² = (7x)² 16/25 = (4/5)² ← √(16/25) = 4/5

a = 7x, b = 4/5 = (7x + 4/5)(7x − 4/5)

💡 Fractions can be perfect squares too! √(16/25) = √16/√25 = 4/5. Always check both numerator and denominator separately for square roots.

Q4 (v) Identity 2 → Identity 3

Factorise: x⁴ − 2x²y² + y⁴

1 Recognise Identity 2 with a = x², b = y²: = (x²)² − 2(x²)(y²) + (y²)² = (x² − y²)²

2 Apply Identity 3 to (x² − y²) = (x+y)(x−y): = [(x + y)(x − y)]² = (x + y)²(x − y)²

Q4 (vi) Identity 3 with composite expressions

Factorise: 4(a + b)² − 9(a − b)²

4(a+b)² = [2(a+b)]² and 9(a−b)² = [3(a−b)]² a = 2(a+b), b = 3(a−b)

= [2(a+b) + 3(a−b)][2(a+b) − 3(a−b)] = [2a + 2b + 3a − 3b][2a + 2b − 3a + 3b] = [5a − b][−a + 5b] = (5a − b)(5b − a)

⛔ This is the most algebraically involved question in Q4. Take extra care when expanding the brackets after the identity is applied — sign errors when distributing 2 and −3 into (a+b) and (a−b) are very common.

Question 5 — Factorising Trinomials Using the x² + (a+b)x + ab Identity

The fourth identity — x² + (a+b)x + ab = (x + a)(x + b) — allows you to factorise quadratic trinomials by finding two numbers whose sum equals the coefficient of x and whose product equals the constant term. This is the most widely used factorisation technique in higher classes.

🔍 The Sum-Product Method for Trinomials

Write the trinomial as x² + px + q (identify p and q).
Find two numbers a and b such that: a + b = p (the middle coefficient) AND a × b = q (the constant term).
Write the answer as (x + a)(x + b).
If p is negative and q is positive → both a and b are negative.
If q is negative → one of a, b is positive and the other is negative.

Q5 (i) Identity 4: x² + (a+b)x + ab

Factorise: a² + 10a + 24

Find two numbers with sum = 10 and product = 24

Factor pair of 24	Sum	Match?
1 × 24	25	✗
2 × 12	14	✗
3 × 8	11	✗
4 × 6	10 ✓	✓ a = 4, b = 6

a² + 10a + 24 = a² + (4 + 6)a + (4 × 6) = (a + 4)(a + 6)

Q5 (ii) Identity 4: x² + (a+b)x + ab

Factorise: x² + 9x + 18

Find two numbers with sum = 9 and product = 18

Factor pair of 18	Sum	Match?
1 × 18	19	✗
2 × 9	11	✗
3 × 6	9 ✓	✓ a = 3, b = 6

x² + 9x + 18 = x² + (3 + 6)x + (3 × 6) = (x + 3)(x + 6)

Q5 (iii) Identity 4 (both numbers negative)

Factorise: p² − 10p + 21

Find two numbers with sum = −10 and product = +21

Both numbers must be negative (negative sum, positive product).

Factor pair of 21	Sum	Match?
−1 × −21	−22	✗
−7 × −3	−10 ✓	✓ a = −7, b = −3

p² − 10p + 21 = p² + [(−7) + (−3)]p + (−7)(−3) = [p + (−7)][p + (−3)] = (p − 7)(p − 3)

Q5 (iv) Identity 4 (one positive, one negative)

Factorise: x² − 4x − 32

Find two numbers with sum = −4 and product = −32

One number positive, one negative (negative product). The larger in magnitude is negative (to get negative sum).

Factor pair of −32	Sum	Match?
−32 × 1	−31	✗
−16 × 2	−14	✗
−8 × 4	−4 ✓	✓ a = −8, b = 4

x² − 4x − 32 = x² + [(−8) + 4]x + (−8)(4) = [x + (−8)][x + 4] = (x − 8)(x + 4)

💡 Sign strategy for Identity 4: If the constant term is positive and middle term is negative → both a, b are negative. If the constant term is negative → one of a, b is positive and the other negative (the one with larger absolute value has the same sign as the middle coefficient).

Quick Reference — All Answers at a Glance

Question	Expression	Identity Used	Factorised Form
Q1(i)	a² + 10a + 25	Identity 1	(a + 5)²
Q1(ii)	l² − 16l + 64	Identity 2	(l − 8)²
Q1(iii)	36x² + 96xy + 64y²	Identity 1	(6x + 8y)²
Q1(iv)	25x² + 9y² − 30xy	Identity 2	(5x − 3y)²
Q1(v)	25m² − 40mn + 16n²	Identity 2	(5m − 4n)²
Q1(vi)	81x² − 198xy + 121y²	Identity 2	(9x − 11y)²
Q1(vii)	(x + y)² − 4xy	Expand + Id.2	(x − y)²
Q1(viii)	l⁴ + 4l²m² + 4m⁴	Identity 1	(l² + 2m²)²
Q2(i)	x² − 36	Identity 3	(x + 6)(x − 6)
Q2(ii)	49x² − 25y²	Identity 3	(7x + 5y)(7x − 5y)
Q2(iii)	m² − 121	Identity 3	(m + 11)(m − 11)
Q2(iv)	81 − 64x²	Identity 3	(9 + 8x)(9 − 8x)
Q2(v)	x²y² − 64	Identity 3	(xy + 8)(xy − 8)
Q2(vi)	6x² − 54	HCF + Id.3	6(x + 3)(x − 3)
Q2(vii)	x² − 81	Identity 3	(x + 9)(x − 9)
Q2(viii)	2x − 32x⁵	HCF + Id.3 ×2	2x(1 + 4x²)(1 + 2x)(1 − 2x)
Q2(ix)	81x⁴ − 121x²	HCF + Id.3	x²(9x + 11)(9x − 11)
Q2(x)	p² − 2pq + q² − r²	Id.2 + Id.3	(p − q + r)(p − q − r)
Q2(xi)	(x+y)² − (x−y)²	Identity 3	4xy
Q3(i)–(ix)	Mixed expressions	HCF / Grouping	See cards above
Q4(i)	x⁴ − y⁴	Id.3 twice	(x² + y²)(x + y)(x − y)
Q4(ii)	a⁴ − (b+c)⁴	Id.3 chain	(a²+b²+c²+2bc)(a+b+c)(a−b−c)
Q4(iii)	l² − (m−n)²	Identity 3	(l+m−n)(l−m+n)
Q4(iv)	49x² − 16/25	Identity 3	(7x + 4/5)(7x − 4/5)
Q4(v)	x⁴ − 2x²y² + y⁴	Id.2 + Id.3	(x+y)²(x−y)²
Q4(vi)	4(a+b)² − 9(a−b)²	Identity 3	(5a−b)(5b−a)
Q5(i)	a² + 10a + 24	Identity 4	(a + 4)(a + 6)
Q5(ii)	x² + 9x + 18	Identity 4	(x + 3)(x + 6)
Q5(iii)	p² − 10p + 21	Identity 4	(p − 7)(p − 3)
Q5(iv)	x² − 4x − 32	Identity 4	(x − 8)(x + 4)

Common Mistakes to Avoid

Not reordering terms in Identity 2: In Q1(iv), 25x² + 9y² − 30xy must be mentally reordered to 25x² − 30xy + 9y² before the identity is visible. Always put the a², −2ab, b² terms in order.
Stopping too early in multi-level problems: In Q2(viii) and Q4(i), after the first application of Identity 3 you get another difference of squares — which must be factorised again. Always check if any factor can be further factorised.
Forgetting the common factor before applying an identity: In Q2(vi) and Q2(ix), the common factor must be taken out first. Applying Identity 3 directly to 6x² − 54 without taking out 6 is incorrect.
Sign errors in Identity 4: When the constant term is negative (Q5 iv), students often get the signs of a and b wrong. Use the factor-pair table method — always list factor pairs of the constant with their signs and check both sum and product.
Confusing Identity 1 and Identity 2: Identity 1 applies when the middle term is positive (+ 2ab). Identity 2 applies when it is negative (− 2ab). The only difference is the sign of the middle term — get this wrong and the entire answer is wrong.

⛔ Board exam alert (Telangana, AP, CBSE): For Q4 and Q5, always write the identity you are using as a separate line before applying it — e.g., write "a² − b² = (a+b)(a−b)" next to your solution. Examiners in TS/AP board papers award 1 mark specifically for stating the identity used.

What Exercise 12.2 Prepares You For

The four algebraic identities introduced in this exercise are foundational to all of secondary school mathematics. The difference of squares identity (a² − b²) appears in Class 9 Polynomials and Class 10 Real Numbers. The perfect square identities underpin the entire chapter on algebraic identities in Class 9.

The trinomial factorisation method from Q5 is the direct precursor to solving Class 10 Quadratic Equations by the factorisation method — one of the most frequently tested topics at the board level.

For CBSE, Telangana, and Andhra Pradesh students, Exercise 12.2 provides essential preparation for 3-mark and 4-mark questions in SA-1 and SA-2 examinations. Mastery of all four identities — and the ability to recognise which one to use — is one of the most valuable skills you can develop in Class 8.

📐 Study tip: Write all four identities on a small card and memorise them. For every trinomial you encounter, first check: "Is this a perfect square (Identity 1 or 2)? A difference of squares (Identity 3)? Or a factorable trinomial (Identity 4)?" This decision process, once automatic, makes factorisation very fast in exams.