Looking for accurate SEBA Class 9 Maths Exercise 2.4 (Polynomials) textual solutions for the 2026–27 academic year? Access complete step-by-step solutions covering standard algebraic identities, evaluating numerical products without direct multiplication, factorizing expressions using identities, expanding trinomials, and cubic expansion identities to score top marks in your annual exams.
EXERCISE 2.4
1. Use suitable identities to find the following products:
(i) (x + 4)(x + 10)
Using identity: (x + a)(x + b) = x² + (a + b)x + ab
= x² + (4 + 10)x + (4 × 10)
= x² + 14x + 40
(ii) (x + 8)(x – 10)
Using identity: (x + a)(x + b) = x² + (a + b)x + ab
= x² + [8 + (-10)]x + [8 × (-10)]
= x² – 2x – 80
(iii) (3x + 4)(3x – 5)
Using identity: (x + a)(x + b) = x² + (a + b)x + ab
= (3x)² + 4 + (-5) + [4 × (-5)]
= 9x² + (-1)(3x) – 20
= 9x² – 3x – 20
(iv) (y² + 3/2)(y² – 3/2)
Using identity: (a + b)(a – b) = a² – b²
= (y²)² – (3/2)²
= y⁴ – 9/4
(v) (3 – 2x)(3 + 2x)
Using identity: (a – b)(a + b) = a² – b²
= (3)² – (2x)²
= 9 – 4x²
2. Evaluate the following products without multiplying directly:
(i) 103 × 107
= (100 + 3)(100 + 7)
= (100)² + (3 + 7)(100) + (3 × 7)
= 10000 + (10)(100) + 21
= 10000 + 1000 + 21
= 11021
(ii) 95 × 96
= (100 – 5)(100 – 4)
= (100)² + (-5) + (-4) + [(-5) × (-4)]
= 10000 + (-9)(100) + 20
= 10000 – 900 + 20
= 9120
(iii) 104 × 96
= (100 + 4)(100 – 4)
= (100)² – (4)²
= 10000 – 16
= 9984
3. Factorise the following using appropriate identities:
(i) 9x² + 6xy + y²
= (3x)² + 2(3x)(y) + (y)²
Using a² + 2ab + b² = (a + b)²
= (3x + y)²
= (3x + y)(3x + y)
(ii) 4y² – 4y + 1
= (2y)² – 2(2y)(1) + (1)²
Using a² – 2ab + b² = (a – b)²
= (2y – 1)²
= (2y – 1)(2y – 1)
(iii) x² – y²/100
= (x)² – (y/10)²
Using a² – b² = (a + b)(a – b)
= (x + y/10)(x – y/10)
4. Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)²
Using identity: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(i) (x + 2y + 4z)²
= (x)² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
= x² + 4y² + 16z² + 4xy + 16yz + 8zx
(ii) (2x – y + z)²
= (2x)² + (-y)² + (z)² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
= 4x² + y² + z² – 4xy – 2yz + 4zx
(iii) (-2x + 3y + 2z)²
= (-2x)² + (3y)² + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx
(iv) (3a – 7b – c)²
= (3a)² + (-7b)² + (-c)² + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)
= 9a² + 49b² + c² – 42ab + 14bc – 6ca
(v) (-2x + 5y – 3z)²
= (-2x)² + (5y)² + (-3z)² + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx
(vi) [(1/4)a – (1/2)b + 1]²
= (a/4)² + (-b/2)² + (1)² + 2(a/4)(-b/2) + 2(-b/2)(1) + 2(1)(a/4)
= a²/16 + b²/4 + 1 – ab/4 – b + a/2
5. Factorise:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
= (2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
= (2x + 3y – 4z)²
= (2x + 3y – 4z)(2x + 3y – 4z)
(ii) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
= (-√2x)² + (y)² + (2√2z)² + 2(-√2x)(y) + 2(y)(2√2z) + 2(2√2z)(-√2x)
Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
= (-√2x + y + 2√2z)²
= (-√2x + y + 2√2z)(-√2x + y + 2√2z)
SEBA Class 9 Maths Exercise 2.4 Solutions
6. Write the following cubes in expanded form:
(i) (2x + 1)³
Using (a + b)³ = a³ + b³ + 3ab(a + b)
= (2x)³ + (1)³ + 3(2x)(1)(2x + 1)
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 1 + 12x² + 6x
= 8x³ + 12x² + 6x + 1
(ii) (2a – 3b)³
Using (a – b)³ = a³ – b³ – 3ab(a – b)
= (2a)³ – (3b)³ – 3(2a)(3b)(2a – 3b)
= 8a³ – 27b³ – 18ab(2a – 3b)
= 8a³ – 27b³ – 36a²b + 54ab²
(iii) [(3/2)x + 1]³
Using (a + b)³ = a³ + b³ + 3ab(a + b)
= (3x/2)³ + (1)³ + 3(3x/2)(1)(3x/2 + 1)
= 27x³/8 + 1 + 9x/2(3x/2 + 1)
= 27x³/8 + 27x²/4 + 9x/2 + 1
(iv) [x – (2/3)y]³
Using (a – b)³ = a³ – b³ – 3ab(a – b)
= (x)³ – (2y/3)³ – 3(x)(2y/3)(x – 2y/3)
= x³ – 8y³/27 – 2xy(x – 2y/3)
= x³ – 8y³/27 – 2x²y + 4xy²/3
7. Evaluate the following using suitable identities:
(i) (99)³
= (100 – 1)³
= (100)³ – (1)³ – 3(100)(1)(100 – 1)
= 1000000 – 1 – 300(99)
= 1000000 – 1 – 29700
= 970299
(ii) (102)³
= (100 + 2)³
= (100)³ + (2)³ + 3(100)(2)(100 + 2)
= 1000000 + 8 + 600(102)
= 1000000 + 8 + 61200
= 1061208
(iii) (998)³
= (1000 – 2)³
= (1000)³ – (2)³ – 3(1000)(2)(1000 – 2)
= 1000000000 – 8 – 6000(998)
= 1000000000 – 8 – 5988000
= 994011992
8. Factorise each of the following:
(i) 8a³ + b³ + 12a²b + 6ab²
= (2a)³ + (b)³ + 3(2a)²(b) + 3(2a)(b)²
Using x³ + y³ + 3x²y + 3xy² = (x + y)³
= (2a + b)³
= (2a + b)(2a + b)(2a + b)
(ii) 8a³ – b³ – 12a²b + 6ab²
= (2a)³ – (b)³ – 3(2a)²(b) + 3(2a)(b)²
Using x³ – y³ – 3x²y + 3xy² = (x – y)³
= (2a – b)³
= (2a – b)(2a – b)(2a – b)
(iii) 27 – 125a³ – 135a + 225a²
= (3)³ – (5a)³ – 3(3)²(5a) + 3(3)(5a)²
Using x³ – y³ – 3x²y + 3xy² = (x – y)³
= (3 – 5a)³
= (3 – 5a)(3 – 5a)(3 – 5a)
(iv) 64a³ – 27b³ – 144a²b + 108ab²
= (4a)³ – (3b)³ – 3(4a)²(3b) + 3(4a)(3b)²
Using x³ – y³ – 3x²y + 3xy² = (x – y)³
= (4a – 3b)³
= (4a – 3b)(4a – 3b)(4a – 3b)
(v) 27p³ – 1/216 – (9/2)p² + (1/4)p
= (3p)³ – (1/6)³ – 3(3p)²(1/6) + 3(3p)(1/6)²
Using x³ – y³ – 3x²y + 3xy² = (x – y)³
= (3p – 1/6)³
= (3p – 1/6)(3p – 1/6)(3p – 1/6)
9. Verify:
(i) x³ + y³ = (x + y)(x² – xy + y²)
R.H.S. = (x + y)(x² – xy + y²)
= x(x² – xy + y²) + y(x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³
= L.H.S.
Hence verified.
(ii) x³ – y³ = (x – y)(x² + xy + y²)
R.H.S. = (x – y)(x² + xy + y²)
= x(x² + xy + y²) – y(x² + xy + y²)
= x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³
= L.H.S.
Hence verified.
Class 9 Maths Chapter 2 Exercise 2.4 Question Answer SEBA 2026-27
10. Factorise each of the following:
(i) 27y³ + 125z³
(ii) 64m³ – 343n³
[Hint: See Question 9.]
(i) 27y³ + 125z³
= (3y)³ + (5z)³
Using a³ + b³ = (a + b)(a² – ab + b²)
= (3y + 5z)[(3y)² – (3y)(5z) + (5z)²]
= (3y + 5z)(9y² – 15yz + 25z²)
(ii) 64m³ – 343n³
= (4m)³ – (7n)³
Using a³ – b³ = (a – b)(a² + ab + b²)
= (4m – 7n)[(4m)² + (4m)(7n) + (7n)²]
= (4m – 7n)(16m² + 28mn + 49n²)
11. Factorise: 27x³ + y³ + z³ – 9xyz
27x³ + y³ + z³ – 9xyz
= (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
Using a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
= (3x + y + z)[(3x)² + y² + z² – (3x)(y) – (y)(z) – (z)(3x)]
= (3x + y + z)(9x² + y² + z² – 3xy – yz – 3zx)
12. Verify that x³ + y³ + z³ – 3xyz = (1/2)(x + y + z)[(x – y)² + (y – z)² + (z – x)²]
R.H.S. = 1/2(x + y + z)[(x – y)² + (y – z)² + (z – x)²]
= 1/2(x + y + z)[(x² – 2xy + y²) + (y² – 2yz + z²) + (z² – 2zx + x²)]
= 1/2(x + y + z)[2x² + 2y² + 2z² – 2xy – 2yz – 2zx]
= 1/2(x + y + z) × 2[x² + y² + z² – xy – yz – zx]
= (x + y + z)(x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz
= L.H.S.
Hence verified.
13.(i) If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
Given x + y + z = 0 ⇒ x + y = -z
Cubing both sides:
(x + y)³ = (-z)³
x³ + y³ + 3xy(x + y) = -z³
Substituting x + y = -z:
x³ + y³ + 3xy(-z) = -z³
x³ + y³ – 3xyz = -z³
x³ + y³ + z³ = 3xyz
Hence proved.
(ii) If a + b + c = 0, then show that a²(b + c) + b²(c + a) + c²(a + b) + 3abc = 0.
Given a + b + c = 0 ⇒ b + c = -a, c + a = -b, a + b = -c
L.H.S. = a²(b + c) + b²(c + a) + c²(a + b) + 3abc
= a²(-a) + b²(-b) + c²(-c) + 3abc
= -(a³ + b³ + c³) + 3abc
Since a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc:
= -(3abc) + 3abc = 0
Hence proved.
(iii) If 2a – b + c = 0, then show that 4a² – b² + c² + 4ac = 0
Given 2a – b + c = 0 ⇒ 2a + c = b
Squaring both sides:
(2a + c)² = b²
4a² + 4ac + c² = b²
4a² – b² + c² + 4ac = 0
Hence proved.
(iv) If a + b + c = 0, then show that (b + c)²/(bc) + (c + a)²/(ca) + (a + b)²/(ab) = 3
Given a + b + c = 0 ⇒ b + c = -a, c + a = -b, a + b = -c
L.H.S. = (b + c)²/bc + (c + a)²/ca + (a + b)²/ab
= (-a)²/bc + (-b)²/ca + (-c)²/ab
= a²/bc + b²/ca + c²/ab
= (a³ + b³ + c³)/abc
Since a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc:
= 3abc/abc = 3
Hence proved.
(v) If a² + b² + c² – ab – bc – ca = 0, then show that a = b = c
Given a² + b² + c² – ab – bc – ca = 0
Multiplying by 2 on both sides:
2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0
(a² – 2ab + b²) + (b² – 2bc + c²) + (c² – 2ca + a²) = 0
(a – b)² + (b – c)² + (c – a)² = 0
Since sum of squares is 0, each term must be 0:
a – b = 0 ⇒ a = b
b – c = 0 ⇒ b = c
c – a = 0 ⇒ c = a
a = b = c
Hence proved.
Algebraic Identities Class 9 SEBA Ex 2.4 Solutions
14. Without actually calculating the cubes, find the value of each of the following:
(i) (-12)³ + (7)³ + (5)³
Let a = -12, b = 7, c = 5
a + b + c = -12 + 7 + 5 = 0
If a + b + c = 0, then a³ + b³ + c³ = 3abc
(-12)³ + (7)³ + (5)³ = 3(-12)(7)(5)
= -1260
(ii) (28)³ + (-15)³ + (-13)³
Let a = 28, b = -15, c = -13
a + b + c = 28 + (-15) + (-13) = 0
If a + b + c = 0, then a³ + b³ + c³ = 3abc
(28)³ + (-15)³ + (-13)³ = 3(28)(-15)(-13)
= 16380
15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a² – 35a + 12
Area = 25a² – 35a + 12
= 25a² – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
= (5a – 4)(5a – 3)
Length = 5a – 4, Breadth = 5a – 3 (or vice versa)
(ii) Area : 35y² + 13y – 12
Area = 35y² + 13y – 12
= 35y² + 28y – 15y – 12
= 7y(5y + 4) – 3(5y + 4)
= (5y + 4)(7y – 3)
Length = 5y + 4, Breadth = 7y – 3 (or vice versa)
SEBA Class 9 Maths Chapter 2 Exercise 2.4 Textbook Answers
16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x² – 12x
Volume = 3x² – 12x
= 3 × x × (x – 4)
Dimensions = 3, x, (x – 4)
(ii) Volume : 12ky² + 8ky – 20k
Volume = 12ky² + 8ky – 20k
= 4k(3y² + 2y – 5)
= 4k(3y² + 5y – 3y – 5)
= 4k[y(3y + 5) – 1(3y + 5)]
= 4k(3y + 5)(y – 1)
Dimensions = 4k, (3y + 5), (y – 1)
Class 9 Maths Exercise 2.4 Expand and Factorize Identities
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