MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.5

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MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.5

Class 9 Maths Chapter 2 Exercise 2.5 MP Board Question 1.
Use suitable identities to find the following products:

  1. (x + 4) (x + 10)
  2. (x + 8) (x – 10)
  3. (3x + 4) (3x – 5)
  4. (y2 + \(\frac{3}{2}\)) (y2 – \(\frac{3}{2}\))
  5. (3 – 2x) (3 + 2x)

Solution:
1. (x + 4)(x + 10)
= (x)2 + (4 + 10)x + 4 x 10
= x2 + 14x + 40

2. (x + 8) (x – 10)
= (x)2 + (8 – 10) x + 8 x – 10
= x2 – 2x – 80

3. (3x + 4) (3x – 5)
= (3x)2 + (4 – 5) (3x) + 4 x (-5)
= 9x2 – 3x – 20

4. (y2 + \(\frac{3}{2}\)) (y2 – \(\frac{3}{2}\))
= (y2)2 + (\(\frac{3}{2}\) – \(\frac{3}{2}\))y + \(\frac{3}{2}\) x \(\frac{-3}{2}\)
= y4 – \(\frac{9}{4}\)

5. (3 – 2x) (3 + 2x)
= (3)2 – (2x)2
= 9 – 4x2

MP Board Solutions

‘Expanding Binomial Calculator‘ is an online tool that helps to calculate the expansion of the given binomial term.

Class 9 Maths Chapter 2 Exercise 2.5 MP Board Question 2.
Evaluate the following products without multiplying directly:

  1. 103 x 107
  2. 95 x 96
  3. 104 x 96

Solution:
1. 103 x 107
= (100 + 3) (100 + 7)
= (100)2 + (3 + 7) x 100 + 3 x 7
[Using identity (x + a) (x + b) = x2 + (a + b) x + ab]
= 10000 + 1000 + 21
= 11021

2. 95 x 96
= (100 – 5) (100 – 4)
[Using identity (a – b)2 = a2 + b2 – 2ab]
= (100)2 + (- 5 – 4) x 100 + (- 5) x (- 4)
= 10000 – 900 + 20
= 9120

3. 104 x 96
= (100 + 4) (100 – 4)
= (100)2 – (4)2
[Using identity (a + b) (a – b) – a2 – b2]
= 10000 – 16
= 9984

MP Board Class 9th Maths Solutions Question 3.
Factorize the following using appropriate identities:

  1. 9x2 + 6xy + y2
  2. 4y2 – 4y + 1
  3. x2 – \(\frac{y^{2}}{100}\)

Solution:
1. 9x2 + 6xy + y2
= 9x2 + 6xy + y2
= (3x)2 + 2 x 3x × y + (y)2
= (3x + y)2
= (3x + y) (3x + y)

2. 4y2 – 4y + 1
= (2y)2 – 2 x 2y x 1 + (1)2
= (2y – 1)2
= (2y – 1) (2y – 1)

3. x2 – \(\frac{y^{2}}{100}\)
= (x)2 – (\(\frac{y^{2}}{10}\))
= (x + \(\frac{y}{10}\))(x – \(\frac{y}{10}\))

MP Board Class 9 Maths Chapter 2 Question 4.
Expand each of the following using suitable identities:

  1. (x + 2y + 4z)2
  2. (2x – y + z)2
  3. (- 2x + 3y + 2z)2
  4. (3a – 7b – c)2
  5. (- 2 + 5y – 3z)2
  6. (\(\frac{1}{4}\)a – \(\frac{1}{2}\)b + 1)

Solution:
1. (x + 2y + 4z)2
= (x)2 + (2y)2 + (4z)2 + 2x(2y) + 2(2y)(4z) + 2 (4z)(x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

2. (2x – y + z)v
= (2x)2 + (-y)2 + (z)2 + 2(2x)(- y) + 2(- y)(z) + 2(z)(2x)
= 4x2 + y2 + z2 – 4xy – 2yz + 4zx

3. (- 2x + 3y + 2z)2
= (-2x)2 + (3y)2 + (2z)2 + 2(- 2x) (3y) + 2(3y)(2z) + 2 (2z) (- 2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12zy – 8zx

4. (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

5. (-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

6. (\(\frac{1}{4}\)a – \(\frac{1}{2}\)b + 1)
MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.5 img-1

Class 9 Maths Chapter 2 Exercise 2.5 Question 5.
Factorize:

  1. 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16 xz
  2. 2x2 +y2 + 8z2 – 2√2xy + 4√2yz – 8xz

Solution:
1. 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (- 4z)2 + 2(2x)(3y) + 2 (3y)(- 4z) + 2 (- 4z) (2x)
= (2x + 3y – 4z)2
= (2x + 3y – 4z) (2x + 3y – 4z)

2. 2x2 + y2 + 8z2 – 2√2xy + 4√2 yz – 8xz
= (- √2 x)2 + (y)2 + (2√2z)2 + 2x (-√2x) (y) + 2(y) (2√2z) + 2 (2√2z ) (-√2x ) = (-√2 x + y + 2√2z)2
= (-√2x + y + 2√2z) (-√2 x + y + 2√2 z)

MP Board Solutions

Class 9 Maths Chapter 2 MP Board Question 6.
Write the following cubes in expanded form:

  1. (2x + 1)3
  2. (2a – 3b)3
  3. (\(\frac{3}{2}\)x + 1)3
  4. (x – \(\frac{2}{3}\)y)3

Solution:
1. (2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 1 + 12x3 + 6x

2. (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b)
= 8a3 – 27b3 – 18ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2

3. (\(\frac{3}{2}\)x + 1)3
MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.5 img-2

4. (x – \(\frac{2}{3}\)y)3
MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.5 img-3

MP Board Class 9th Maths Chapter 2 Question 7.
Evaluate the following using suitable identities:

  1. (99)3
  2. (102)3
  3. (998)3

Solution:
1. (99)3 = (100 – 1)3
= (100)3 – (1)3 – 3(100)(1)(100 – 1)
= 1000000 – 1 – 300 (100 – 1)
= 1000000 – 1 – 300 x 99
= 999999 – 29700 = 970299

2. (102)3 = (100 + 2)3
= (100)3 + (2)3 + 3 (100)(2)(100 + 2)
= 1000000 + 8 + 600 (100 + 2)
= 1000008 + 61200
= 1061208

3. (998)3 = (1000 – 2)3
= (1000)3 – (2)3 – 3(1000)(2)(1000 – 2)
= 1000000000 – 8 – 6000 x 998
= 994011992

MP Board Solution Class 9th Maths Question 8.
Factorize each of the following:

  1. 8a3 + b3 + 12a2b + 6ab2
  2. 8a3 – b3 – 12a2b + 6ab2
  3. 27 – 125a3 – 135a + 225a2
  4. 64a3 – 27b3 – 144a2b + 108ab2
  5. 27p3 – \(\frac{1}{216}\) – \(\frac{9}{2}\)p2 + \(\frac{1}{4}\)p

Solution:
1. 8a3 + b3 + 12a2b + 6ab2
= (2a)3 + (b)3 + 3(2a)(b)(2a + b)
= (2a + b)3
= (2a + b)(2a + b)(2a + b)

2. 8a3 – b3 – 12a2b + 6ab2
= (2a)3 – (b)3 – 3(2a)(b)(2a – b)
= (2a – b)3
= (2a – b) (2a – b) (2a – b)

3. 27 – 125a3 – 135a + 225a2
= (3)3 – (5a)3 – 3(3)(5a)(3 – 5a)
= (3 – 5a)3
= (3 – 5a)(3 – 5a)(3 – 5a)

4. 64a3 – 27b3 – 144a2b + 108ab2
= (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)
= (4a – 3b)3
= (4a – 3b)(4a – 3b)(4a – 3b)

5. 27p3 – \(\frac{1}{216}\) – \(\frac{9}{2}\)p2 + \(\frac{1}{4}\)p
MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.5 img-4

9th Class Maths Exercise 2.5 Question 9.
Verify:

  1. x3 + y3 = (x + y) (x2 – xy + y2)
  2. x3 – y = (x – y) (x2 + xy + y2)

Solution:
1. x3 + y3 = (x + y) (x2 – xy + y2)
RHS = (x + y) (x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= x2 – x2y + xy2 + x2y – xy2 + y3
= x3 + y2 = LHS

2. x3 – y3 = (x – y)(x2 + xy + y2)
RHS = (x – y) (x2 + xy + y2)
= x(x2 + xy + y2) – y(x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3 = LHS .

MP Board Solutions

MP Board 9th Maths Solutions Question 10.
Factorize each of the following:

  1. 27y3 + 125z3
  2. 64m3 – 343n2

Solution:
1. 21y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z) [(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z) (9y2 – 15yz + 25z2)

2. 64m3 – 343n3 = (4m)3 – (7n)3
= (4m – 7n) [(4m)2 + (4m)(7n) + (7n)2]
= (4m – In) (16m2 + 28mn + 49n2)

Class 9th Chapter 2.5 Question 11.
Factorize: 27x3 + y3 + z3 – 9xyz
Solution:
27x3 + y3 + z3 – 9xyz
= (3x)3 + (y)3 + (z)3 – 3 x 3x × y x z
= (3x + y + z) {(3x)2 + (y)2 + (z)2 – 3xy – yz – 3xz}
= (3x + y + z) (9x2 + y2 + z2 – 3xy – yz – 3xz)

9th Class Maths 2.5 Question 12.
Verify that x3 + y3 + z3 – 3xyz = \(\frac { 1 }{ 2 } \) (x + y + z)[(x-y)2 + (y – z)2 + (z – x)2]
Solution:
x3 + y3 + z3 – 3xyz
= \(\frac { 1 }{ 2 } \) (x + y + z) [(x – y)2 + (y – z)2 (z – x)2]
RHS = \(\frac { 1 }{ 2 } \) (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
= \(\frac { 1 }{ 2 } \) (x + y + 2) [(x2 + 2xy + y2) + (y2 – 2yz + z2) + (z2 – 2zx + x2)]
= \(\frac { 1 }{ 2 } \) (x + y + z) [2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx]
= \(\frac { 1 }{ 2 } \) (x + y + z) x 2(x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = LHS

9th Class Math Exercise 2.5 Question 13.
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz
Solution:
We know x3 + y3 + z3 – 3xyz
= (x + y + z) (x2 + y2 + z2 – xy – yz – xz)
= x3 + y3 + z3 – 3xyz
= 0 x (x2 + y2 + z2 – xy – yz – xz) (∴ x + y + z = 0)
x3 + y3 + z3 – 3xyz = 0
x3 + y3 + z3 = 3xyz

Class 9 Maths 2.5 In Hindi Question 14.
Without actually calculating the cubes, find the value of each of the following:

  1. (-12)3 + (7)3 + (5)3
  2. (28)3 + (-15)3 ÷ (-13)3

Solution:
1. (-12)3 + (7)3 + (5)3
a + b + c = – 12 + 7 + 5
= – 12 + 12 = 0
(- 12)3 + (7)3 + (5)3
= 3 (- 12) (7) (5) = – 1260

2. (28)3 + (- 15)3 + (- 13)3
a + b + c = 28 + (- 15) + (- 13)
= 28 – 28 = 0
∴ (28)3 + (-15)3 + (- 13)3
= 3(28) (- 15) (- 13) = 16380

Class 9th Maths Chapter No 2 Exercise 2.5 Question 15.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

  1. Area: 25a2 – 35a + 12 …..(i)
  2. Area: 35y2 + 13y – 12 …..(ii)

Solution:
1. A = 25a2 – 35a + 12
= 25a2 – 20a – 15a + 12
= 5a (5a – 4) – 3 (5a – 4)
= (5a – 4) (5a – 3)
Length = (5a – 4) and breadth = 5a – 3

2. A = 35y2 + 13y – 12
= 35y2 + 28y – 15y – 12
= 7y (5y + 4) – 3 (5y + 4)
= (7y – 3) (5y + 4)
Length = (7y – 3) and breadth = (5y + 4)

Note: In the above questions out of the two factors, any one factor can be taken as length and the other factor will be breadth.

MP Board Solutions

Exercise 2.5 Class 9 Question 16.
What are the possible expressions for the dimensions of the cuboids, whose volumes are given below:

  1. Volume: 3x2 – 12x …(i)
  2. Volume: 12ky2 + 8ky – 20k …(ii)

Solution:
1. V = 3x2 – 12x
= 3x (x – 4) = 3 × x × (x – 4)
l = 3
b = x
h = (x – 4)

2. V = 12ky2 + 8ky – 20k
= 4k (3y2 + 2y – 5)
= 4k (3y2 + 5y – 3y – 5)
= 4k [y (3y + 5) – 1 (3y + 5)]
= 4k (3y + 5) (y – 1)
l = 4 k
b = (3y + 5)
h = (y – 1)

MP Board Class 9th Maths Solutions

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