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

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

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

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

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 – \(\frac{y}{10}\))

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

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

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

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

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

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

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)

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)

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

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

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

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

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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