Toppers Study Toppers Study Toppers StudyDetailed NCERT Solutions for 9 Mathematics 2. Polynomials to simplify learning. Understand chapters clearly and practice with free solutions for better results.
Detailed NCERT Solutions for 9 Mathematics 2. Polynomials to simplify learning. Understand chapters clearly and practice with free solutions for better results.
Preparing for exams becomes easier with Exercise 2.5. Whether you are studying for board exams or mid-term exams, 9 Mathematics Chapter 2. Polynomials solutions provide quick revising points, well-structured answers, and additional practice material to help you score better.
ncert_solutionsAlgebraic Identities:
∵ ∴
(1) (x + y)2 = x2 + 2xy + y2
(2) (x - y)2 = x2 - 2xy + y2
(3) x2 - y2 = (x + y) (x - y)
(4) (x + a) (x + b) = x2 + (a + b)x + ab
(5) (x + y)3 = x3 + 3x2y + 3xy2 + y3
(6) (x - y)3 = x3 - 3x2y + 3xy2 - y3
(7) x3 + y3 = (x + y) (x2 - xy + y2)
(8) x3 - y3 = (x - y) (x2 + xy + y2)
(9) (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
(10) x3 + y3 + z3 - 3xyz = ( x + y + z) (x2 + y2 + z2 - xy - yz - zx)
Exercise 2.5
Q1. Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(v) (3 – 2x) (3 + 2x)
Solution:
(i) (x + 4) (x + 10)
Using identity; (x + a) (x + b) = x2 + (a + b)x + ab
(x + 4) (x + 10) = 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
(x + 8) (x – 10) = 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 + 4) (3x – 5) = (3x)2 + [4 + (-5)]3x + (4)(-5)
= 9x2 - 3x - 20
Using identity; (x + y) (x - y) = x2 - y2
(v) (3 – 2x) (3 + 2x)
Using identity; (x + y) (x - y) = x2 - y2
(3 – 2x) (3 + 2x) = (3)2 - (2x)2
= 9 - 4x2
Q2. Evaluate the following products without multiplying directly:
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
Solution:
(i) 103 × 107 = (100 + 3) (100 + 7)
Using identity; (x + a) (x + b) = x2 + (a + b)x + ab
(100 + 3) (100 + 7) = (100)2 + (3 + 7)100 + 3×7
=10000 + 1000 + 21
= 11021
(ii) 95 × 96 = (90 + 5) (90 + 6)
Using identity; (x + a) (x + b) = x2 + (a + b)x + ab
(90 + 5) (90 + 6) = (90)2 + (5 + 6)90 + 5×6
=8100 + 990 + 30
= 9120
(iii) 104 × 96 = (100 + 4) (100 - 4)
Using identity; (x + y) (x - y) = x2 - y2
(100)2 - (4)2
=10000 - 16
= 9984
3. Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2
(ii) 4y2 – 4y + 1
Solution:
(i) 9x2 + 6xy + y2
= (3x)2 + 2.3x.y + (y)2 [ ∵ x2 + 2xy + y2 = (x + y)2]
∴ = (3x + y)2
= (3x + y) (3x + y)
(ii) 4y2 - 4y + 1
= (2y)2 - 2.2y.1 + (1)2 [ ∵ x2 - 2xy + y2 = (x - y)2]
∴ = (2y - 1)2
= (2y - 1) (2y - 1)
[ ∵ x2 - y2 = (x + y) (x - y) ]
Q4. Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)2
(ii) (2x – y + z)2
(iii) (–2x + 3y + 2z)2
(iv) (3a – 7b – c)2
(v) (–2x + 5y – 3z)2
Solution:
(i) (x + 2y + 4z)2
Here let as a = x, b = 2y, c = 4z and putting the values of a, b and c in the
Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (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
Here let as a = 2x, b = - y, c = z and putting the values of a, b and c in the
Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (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
Here let as a = - 2x, b = 3y, c = 2z and putting the values of a, b and c in the
Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (–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
Here let as x = 3a, y = – 7b, z = – c and putting the values of x, y and z in the
Identity (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
∴ (3a – 7b – c)2
= (3a)2 + (– 7b)2 + (– c)2 + 2(3a)(– 7b) + 2(– 7b)(– c) + 2(– c)(3a)
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ac
(v) (–2x + 5y – 3z)2
Here let as a = - 2x, b = 5y, c = –3z and putting the values of a, b and c in the
Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (–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
Q5. Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Solution:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (4z)2 + 2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x)
[∵ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2 ]
= (2x + 3y + 4z)2
= (2x + 3y + 4z) (2x + 3y + 4z)
Q6. Write the following cubes in expanded form:
(i) (2x + 1)3
(ii) (2a – 3b)3
Solution:
(i) (2x + 1)3
[using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]
(2x + 1)3 = (2x)3 + 3(2x)2(1) + 3(2x)(1)2 + (1)3
= 8x3 + 12x2 + 6x + 1
(ii) (2a – 3b)3
[Using identity (x - y)3 = x3 - 3x2y + 3xy2 - y3]
(2a - 3b)3 = (2a)3 - 3(2a)2(3b) + 3(2a)(3b)2 - (3b)3
= 8a3 - 36a2b + 54ab2 - 27b3
[using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]
[using identity (a - b)3 = a3 - 3a2b + 3ab2 - b3]
Q7. Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) (99)3
= (100 - 1)3
[using identity (a - b)3 = a3 - 3a2b + 3ab2 - b3]
(100 - 1)3 = (100)3 - 3(100)2(1) + 3(100)(1)2 - (1)3
= 1000000 - 30000 + 300 - 1
= 1000300 - 30001
= 970299
(ii) (102)3
= (100 + 2)3
[using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]
(100 + 2)3 = (100)3 + 3(100)2(2)+ 3(100)(2)2 + (2)3
= 1000000 + 60000 + 1200 + 8
= 1061208
(iii) (998)3
= (1000 - 2)3
[using identity (a - b)3 = a3 - 3a2b + 3ab2 - b3]
(1000 - 2)3 = (1000)3 - 3(1000)2(2)+ 3(1000)(2)2 - (2)3
= 1000000000 - 6000000 + 12000 - 8
= 1000012000 - 6000008
= 994011992
Q8. Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a2 – b2 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a2
(iv) 64a3 – 27b3 – 144a2b + 108ab2
Solution:
(i) 8a3 + b3 + 12a2b + 6ab2
= (2a)3 +(b)3 + 3(2a)2(b) + 3(2a)(b)2
[Using identity x3 + y3 + 3x2y + 3xy2 = (x + y)3 ]
= (2a)3 +(b)3 + 3(2a)2(b) + 3(2a)(b)2 = (2a + b)3
= (2a + b)(2a + b)(2a + b)
(ii) 8a2 – b2 – 12a2b + 6ab2
= (2a)3 - (b)3 - 3(2a)2(b) + 3(2a)(b)2
[Using identity x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]
= (2a)3 - (b)3 - 3(2a)2(b) + 3(2a)(b)2 = (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 identity x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]
= (3)3 - (5a)3 - 3(3)2(5a) + 3(3)(5a)2= (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 identity x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]
= (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a)(3b)2 = (4a - 3b)3
= (4a - 3b)(4a - 3b)(4a - 3b)
[Using identity x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]
Q9. Verify:
(i) x3 + y3 = (x + y) (x2 – xy + y2)
Solution:
RHS = (x + y) (x2 – xy + y2)
= x(x2 – xy + y2) + y (x2 – xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 + y3
∵ LHS = RHS Verified
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
RHS = (x - y) (x2 + xy + y2)
x(x2 + xy + y2) - y(x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3
∵ LHS = RHS Verified
Q10. Factorise each of the following:
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
Solution:
(i) 27y3 + 125z3
= (3y)3 + (5z)3
[Using identity x3 + y3 = (x + y) (x2 – xy + y2) ]
(3y)3 + (5z)3 = (3y + 5y) [(3y)2 - (3y)(5z) + (5z)2]
= (3y + 5y) (9y2 - 15yz + 25z2)
(ii) 64m3 – 343n3
Solution:
(ii) 64m3 – 343n3
= (4m)3 – (7n)3
[Using identity x3 – y3 = (x – y) (x2 + xy + y2) ]
(4m)3 – (7n)3 = (4m – 7n) [(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n) (16m2 + 28mn + 49n2)
Q11. Factorise : 27x3 + y3 + z3 – 9xyz
Solution:
= (3x)3 + (y)3 + (z)3 - 9xyz
∵ x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - zx)
Using identity:
= (3x + y + z) ((3x)2 + (y)2 + (z)2 - (3x)(y) - (y)(z) - (z)(3x))
= (3x + y + z) (9x2 + y2 + z2 - 3xy - yz - 3zx)
Q12. Verify that:
x3 + y3 + z3 - 3xyz =½ (x + y + z) [(x -y)2 + (y - z)2 + (z - x)2]
LHS = ½(x + y + z) [x2 - 2xy + y2 + y2 - 2yz + z2 + z2 - 2xz + x2]
= ½(x + y + z) (2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz)
= ½ × 2(x + y + z)(x2 + y2 + z2 - xy - yz - xz)
= (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
= x3 + y3 + z3 - 3xyz [Using Identity]
LHS = RHS
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