Study Materials

# NCERT Solutions for Class 9th Mathematics

Page 5 of 5

## Chapter 2. Polynomials

### Exercise 2.5

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 + z- 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 + c 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)= (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)= (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)= (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  + y+ 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 - y- 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 - y- 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 - y- 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 + 49n​2)

Q11. Factorise : 27x3 + y3 + z3 – 9xyz

Solution:

= (3x)3 + (y)3 + (z)- 9xyz

∵ x+ y3 + z3 - 3xyz =  (x + y + z) (x2 + y2 + z​2 - 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:

x+ 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) (2x+ 2y2 + 2z2 - 2xy - 2yz - 2xz)

= ½ × 2(x + y + z)(x+ y2 + z2 - xy - yz - xz)

= (x + y + z)(x+ y2 + z2 - xy - yz - xz

= x+ y3 + z3 - 3xyz                 [Using Identity]

LHS = RHS

Page 5 of 5

Chapter Contents: