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Exercise 2.5 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study
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Exercise 2.5 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study
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Exercise 2.5 class 9 Mathematics Chapter 2. Polynomials
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2. Polynomials
| Exercise 2.5 |
Exercise 2.5 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study
Algebraic 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
Other Pages of this Chapter: 2. Polynomials
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