Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 Textbook Exercise Questions and Answers.

## RBSE Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.5

Question 1.

Use suitable identities to find the following products :

(i) (x + 4) (x + 10)

Answer:

(x + 4)(x + 10) = x^{2} + (4 + 10)x + 4 × 10 = x^{2} + 14x + 40

(ii) (x + 8) (x – 10)

Answer:

(x + 8)(x – 10) = x^{2} + (8 – 10)x + 8 × (- 10) = x^{2} – 2x – 80

(iii) (3x + 4) (3x – 5)

Answer:

(3x + 4)(3x – 5) = (3x)^{2} + (4 – 5)3x + 4 x (- 5)

[Using (x + a)(x + b) = x^{2} + (a + b)x + ab. Here, x is taken = 9x^{2} – 3x – 20

(iv) (y^{2} + \(\frac{3}{2}\))(y^{2} – \(\frac{3}{2}\))

Answer:

(y^{2} + \(\frac{3}{2}\))(y^{2} – \(\frac{3}{2}\)) = (y^{2})^{2} – (\(\frac{3}{2}\))^{2} = y^{4} – \(\frac{9}{4}\)

(v) (3 – 2x)(3 + 2x)

Answer:

(3 – 2x)(3 + 2x) = (3)^{2} – (2x)^{2} = 9 – 4x^{2}

Question 2.

Evaluate the following products without multiplying directly :

(i) 103 × 107

Answer:

103 × 107 = (100 + 3)(100 + 7) = (100)^{2} + (3 + 7)(100) + 3 × 7

= 100 × 100 + (10)(100) + 21 = 10000 + 1000 + 21 = 11021

(ii) 95 × 96

Answer:

95 × 96 = (100 – 5)(100 – 4) = (100)^{2} + (- 5 – 4)(100) + (- 5)(- 4)

= 100 × 100 + (- 9)(100) + 20

= 10000 – 900 + 20 = 9120

(iii) 104 × 96

Answer:

104 × 96 = (100 + 4) (100 – 4) = (100)^{2} – (4)^{2}

= 10000 – 16 = 9984

Question 3.

Factorise the following using appropriate identities :

(i) 9x^{2} + 6xy + y^{2}

Answer:

9x^{2} + 6xy + y^{2} = (3x)^{2} + 2(3x)(y) + (y)^{2}

= (3x + y)^{2} = (3x + y)(3x + y)

(ii) 4y^{2} – 4y + 1

Answer:

4y^{2} – 4y + 1 = (2y)^{2} – 2(2y)(1) + (1)^{2}

= (2y – 1)^{2} = (2y – 1)(2y – 1)

(iii) x^{2} – \(\frac{y^{2}}{100}\)

Answer:

x^{2} – \(\frac{y^{2}}{100}\) = x^{2} – (\(\frac{y}{10}\))^{2}

= (x – \(\frac{y}{10}\))(x – \(\frac{y}{10}\))

Question 4.

Expand each of the following, using suitable identities :

(i) (x + 2y + 4z)^{2}

Answer:

(x + 2y + 4z)^{2} = x^{2} + (2y)^{2} + (4z)^{2} + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

= x^{2} + 4y^{2} + 16z^{2} + 4xy + 16yz + 8zx

(ii) (2x – y + z)^{2}

Answer:

(2x – y + z)^{2} = [2x + (- y) + z]^{2}

= (2x)^{2} + (-y)^{2} + z^{2} + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)

= 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4zx

(iii) (- 2x + 3y + 2z)^{2}

Answer:

(- 2x + 3y + 2z)^{2} = [(- 2x) + 3;y + 2z]^{2}

= (- 2x)^{2} + (3y)^{2} + (2z)^{2} + 2(- 2x)(3y) + 2(3y)(2z) + 2(2z)(- 2x)

= 4x^{2} + 9y^{2} + ^{2} – 12xy + 12yz – 8zx

(iv) (3a – 7b – c)^{2}

Answer:

(3a – 76 – c) = [3a + (- 7b) + (- c)]^{2}

= (3a)^{2} + (- 7b)^{2} + (- c)^{2} + 2(3a)(- 76) + 2(- 7b)(- c) + 2(- c)(3a)

= 9a^{2} + 49b^{2} + c^{2} – 42a6 + 146c – 6ca

(v) (- 2x + 5y – 3z)^{2}

Answer:

(- 2x + 5y – 3z)^{2} = [(- 2x) + 5y + (- 3z)]^{2}

= (- 2x)^{2} + (5y)^{2} + (- 3z)^{2} + 2(- 2x)(5y) + 2(5y)(- 3z) + 2(-3z)(-2x)

= 4x^{2} + 25y^{2} + 9z^{2} – 20xy – 30yz + 12zx

(vi) [\(\frac{1}{4}\)a – \(\frac{1}{2}\)b + 1]^{2}

Answer:

Question 5.

Factorise :

(i) 4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

Answer:

4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

= (2x)^{2} + (3y)^{2} + (- 4z)^{2} + 2(2x)(3y) + 2(3y)(- 4z) + 2(2x)(- 4z)

= [2x + 3y + (- 4z)]^{2} = (2x + 3y- 4z) = (2x + 3y- 4z)(2x + 3y- 4z)

(ii) 2x^{2} + y^{2} + 8z^{2} – 2√2xy + 4√2yz – 8xz

Answer:

2x^{2} + y^{2} + 8z^{2} – 2√2xy + 4√2 yz – 8xz

= (- √2x)^{2} + y^{2} + (2√2z)^{2} + 2(- y√2 x)(y) + 2(y){2√2z) + 2(-x)(2√2z)

= (-√2x + y + 2√2z)^{2} = (-√2x + y + 2√2 z)(-√2 x + y + 2√2 z)

Question 6.

Write the following cubes in expanded form :

(i) (2x + 1)^{3}

Answer:

(2x + 1)^{3} = (2x)^{3} + 3(2x)^{2}(1) + 3(2x)(1)^{2} + (1)^{3}

= 8x^{3} + 12x^{2} + 6x + 1

(ii) (2a – 3b)^{3}

Answer:

(2a – 3b)^{3} = (2a)^{3} – 3(2a)^{2}(3b) + 3(2a)(3b)^{2} – (3b)^{3}

= 8a^{3} – 36a^{2}b + 54ab^{2} – 27b^{3}

(iii) [\(\frac{3}{2}\)x + 1]^{3}

Answer:

(iv) [x – \(\frac{2}{3}\)y]^{3}

Answer:

[x – \(\frac{2}{3}\)y]^{3} = x^{3} – 3(x)^{2}(\(\frac{2}{3}\))^{2} – (\(\frac{2}{3}\)y)^{3}

= x^{3} – 2x^{2}y + \(\frac{4}{3}\)xy^{2} – \(\frac{8}{27}\)y^{3}

Question 7.

Evaluate the following using suitable identities :

(i) (99)^{3}

Answer:

(99)^{3} = (100 – 1)^{3} = (100)^{3} – 1^{3} – 3(100) (1) (100 – 1)

= 1000000 – 1 – 29700

= 970299

(ii) (102)^{3}

Answer:

(102)^{3} = (100 + 2)^{3}

= (100)^{3} + (2)^{3} + 3(100) (2) (100 + 2)

= 1000000 + 8 + 61200 = 1061208

(iii) (998)^{3}

Answer:

(998)^{3} = (1000 – 2)^{3}

= (1000)^{3} – (2)^{3} – 3(1000) (2) (1000 – 2)

= 1000000000 – 8 – 5988000 = 994011992

Question 8.

Factorise each of the following :

(i) 8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}

Answer:

8a^{3} + b^{3} + 12a^{2}b + 6ab^{2} = (2a)^{3} + (6)^{3} + 3(2a)(b)(2a + b)

= (2a + b)^{3}

= (2a + b)(2a + b)(2a + b)

(ii) 8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}

Answer:

8a^{3} – b^{3} – 12a2^{2}b + 6ab^{2}

= (2a)^{3} – b^{3} – 3(2a)(6)(2a – b)

= (2a – b)^{3}

= (2a – b)(2a – b)(2a – b)

(iii) 27 – 125a^{3} – 135a + 225a^{2}

Answer:

27 – 125a^{3} – 135a + 225a^{2} = (3)^{3} – (5a)^{3} – 3(3)(5a)(3 – 5a)

= (3 – 5a)^{3}

= (3 – 5a)(3 – 5a)(3 – 5a)

(iv) 64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}

Answer:

64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2} = (4a)^{3} – (3b)^{3} – 3(4a)(3b)(4a – 36)

= (4a – 36 )^{3}

= (4a – 3b)(4a – 3b)(4a – 3b)

(v) 27p^{3} – \(\frac{1}{216}\) – p\(\frac{9}{2}\)^{2} + \(\frac{1}{4}\)p

Answer:

Question 9.

Verify :

(i) x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})

Answer:

RHS = (x + y)(x^{2} – xy + y^{2}) = x(x^{2} – xy + y^{2}) + y(x^{2} – xy + y^{2})

= x^{3} – x^{2}y + xy^{2} + x^{2}y – xy^{2} + y^{3}

– x^{3} + y^{3} = L.H.S

Hence proved.

(ii) x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

Answer:

RHS = (x – y)(x^{2} + xy + y^{2}) = x(x^{2} + xy + y^{2}) – y(x^{2} + xy + y^{2})

= x^{3} + x^{2}y + xy^{2} – x^{2}y – xy^{2} – y^{3}

= x^{3} – y^{3} = LHS Hence proved.

Question 10.

Factorise each of the following :

(i) 27y^{3} + 125z^{3}

Answer:

27y^{3} + 125z^{3} = (3y)^{3} + (5z)^{3} = (3y + 5z)[(3y)^{2} – (3y)(5z) + (5z)^{2}]

= (3y + 5z)(9y^{2} – 15yz + 25z^{2})

(ii) 64m^{3} – 343n^{3} (Hint: See question 9)

Answer:

64m^{3} – 343n^{3} = (4m)^{3} – (7n)^{3} = (4m – 7n)[(4m)^{2} + (4m)(7n) + (7n)^{2}]

= (4m – 7n)(16m^{2} + 28mn + 49n^{2})

Question 11.

Factorise : 27x^{3} + y^{3} + z^{3} – 9xyz

Answer:

27x^{3} + y^{3} + z^{3} – 9xyz= (3x)^{3} + y^{3} + z^{3} – 3(3x)(y)(z)

= (3x + y + z)[(3x)^{2} + y^{2} + z^{2} – (3x)y – yz – z(3x)] = (3x +y +z)(9x^{2} + y^{2} +z^{2} – 3xy -yz – 3zx)

Question 12.

Verify that x^{3} + y^{3} + z^{3} – 3xyz = \(\frac{1}{2}\) (x + y + z)[(x – y)2 + (y – z)2 + (z – x)^{2}].

Answer:

RHS = \(\frac{1}{2}\) (x + y + z)[(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]

= \(\frac{1}{2}\)(x + y+ z) (^{2} – 2xy + y^{2} + y^{2} – 2yz + z^{2} + z^{2} – 2zx + x^{3})

= (x + y + z)(x^{2} + y^{2} + z^{2} – yz – zx – xy)

Hence proved.

Question 13.

If x + y + z = 0, show that x^{3} + y^{3} + z^{3} = 3xyz.

Answer:

We have : x + y + z = 0 x + y = -z

Cubing both sides, we get:

(x + y)^{3} = (- z)^{3}

⇒ x^{3} + y^{3} + 3xy(x + y) = – z^{3}

⇒ x^{3} + y^{3} – 3xyz = -z

⇒ x^{3} + y^{3} + z^{3} – 3xyz. (∵ x + y = – z)

Hence proved.

Question 14.

Without actually calculating the cubes, find the value of each of the following :

(i) (- 12)^{3} + (7)^{3} + (5)^{3}

Answer:

Let x = – 12, y = 7 and z = 5. .

Here, x + y + z =-12 + 7 + 5 = 0

x^{3} + y^{3} + z^{3} = 3xyz

or (- 12)^{3} + (7)^{3} + (5)^{3}

= 3 × (- 12) × 7 × 5 = – 1260

(ii) (28)^{3} + (- 15)^{3} + (- 13)^{3}

Answer:

Let x = 28, y = – 15 and z = – 13.

Here, x + y + z = 28 – 15 – 13 = 0

x^{3} + y6 + z‘ = 3xyz

or (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 :

Answer:

Possible length and breadth of the rectangles are the factors of its given area.

(i) Area = 25a^{2} – 35a + 12 = 25a^{2} – 15a – 20a + 12 .

= 5a(5a – 3) – 4(5a – 3) = (5a – 3)(5a – 4)

∴ Possible expressions for length and breadth are (5a – 3) and (5a – 4) units.

(ii) Area = 35y^{2} + 13y – 12 = 35y^{2} + 28y – 15y – 12

= 7y(5y + 4) – 3(5y + 4) = (5y + 4)(7y – 3)

∴ Possible expressions for length and breadth are (5y + 4) and (7y – 3) units.

Question 16.

What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

Answer:

Possible expressions for the dimensions of the cuboids are the factors of their volumes.

(i) Volume = 3a^{2} – 12a = 3a(a – 4)

∴ Possible expressions for the dimensions of cuboid are 3, a and (x – 4) units.

(ii) Volume = 12ky^{2} + 8ky – 20k = 4k(3y^{2} + 2y – 5)

= 4k(3y^{2} – 3y + 5y – 5) = 4k[3y(y – 1) + 5(y – 1)]

= 4k(y – 1)(3y + 5)

∴ Possible expressions for the dimensions of cuboid are 4k, (y – 1) and (3y + 5) units.

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