RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

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) = x² + (4 + 10)x + 4 × 10 = x² + 14x + 40

(ii) (x + 8) (x – 10)
Answer:
(x + 8)(x – 10) = x² + (8 – 10)x + 8 × (- 10) = x² – 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) = x² + (a + b)x + ab. Here, x is taken = 9x² – 3x – 20

(iv) (y² + \(\frac{3}{2}\))(y² – \(\frac{3}{2}\))
Answer:
(y² + \(\frac{3}{2}\))(y² – \(\frac{3}{2}\)) = (y²)² – (\(\frac{3}{2}\))² = y⁴ – \(\frac{9}{4}\)

(v) (3 – 2x)(3 + 2x)
Answer:
(3 – 2x)(3 + 2x) = (3)² – (2x)² = 9 – 4x²

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) 9x² + 6xy + y²
Answer:
9x² + 6xy + y² = (3x)² + 2(3x)(y) + (y)²
= (3x + y)² = (3x + y)(3x + y)

(ii) 4y² – 4y + 1
Answer:
4y² – 4y + 1 = (2y)² – 2(2y)(1) + (1)²
= (2y – 1)² = (2y – 1)(2y – 1)

(iii) x² – \(\frac{y^{2}}{100}\)
Answer:
x² – \(\frac{y^{2}}{100}\) = x² – (\(\frac{y}{10}\))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 = x² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
= x² + 4y² + 16z² + 4xy + 16yz + 8zx

(ii) (2x – y + z)²
Answer:
(2x – y + z)² = [2x + (- y) + z]²
= (2x)² + (-y)² + z² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
= 4x² + y² + z² – 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)
= 4x² + 9y² + 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)
= 9a² + 49b² + c² – 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)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

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

Question 5.
Factorise :
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
Answer:
4x² + 9y² + 16z² + 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) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
Answer:
2x² + y² + 8z² – 2√2xy + 4√2 yz – 8xz
= (- √2x)² + y² + (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
= 8x³ + 12x² + 6x + 1

(ii) (2a – 3b)³
Answer:
(2a – 3b)³ = (2a)³ – 3(2a)²(3b) + 3(2a)(3b)² – (3b)³
= 8a³ – 36a²b + 54ab² – 27b³

(iii) [\(\frac{3}{2}\)x + 1]³
Answer:

(iv) [x – \(\frac{2}{3}\)y]³
Answer:
[x – \(\frac{2}{3}\)y]³ = x³ – 3(x)²(\(\frac{2}{3}\))² – (\(\frac{2}{3}\)y)³
= x³ – 2x²y + \(\frac{4}{3}\)xy² – \(\frac{8}{27}\)y³

Question 7.
Evaluate the following using suitable identities :
(i) (99)³
Answer:
(99)³ = (100 – 1)³ = (100)³ – 1³ – 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) 8a³ + b³ + 12a²b + 6ab²
Answer:
8a³ + b³ + 12a²b + 6ab² = (2a)³ + (6)³ + 3(2a)(b)(2a + b)
= (2a + b)³
= (2a + b)(2a + b)(2a + b)

(ii) 8a³ – b³ – 12a²b + 6ab²
Answer:
8a³ – b³ – 12a2²b + 6ab²
= (2a)³ – b³ – 3(2a)(6)(2a – b)
= (2a – b)³
= (2a – b)(2a – b)(2a – b)

(iii) 27 – 125a³ – 135a + 225a²
Answer:
27 – 125a³ – 135a + 225a² = (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)
= (3 – 5a)³
= (3 – 5a)(3 – 5a)(3 – 5a)

(iv) 64a³ – 27b³ – 144a²b + 108ab²
Answer:
64a³ – 27b³ – 144a²b + 108ab² = (4a)³ – (3b)³ – 3(4a)(3b)(4a – 36)
= (4a – 36 )³
= (4a – 3b)(4a – 3b)(4a – 3b)

(v) 27p³ – \(\frac{1}{216}\) – p\(\frac{9}{2}\)² + \(\frac{1}{4}\)p
Answer:

Question 9.
Verify :
(i) x³ + y³ = (x + y) (x² – xy + y²)
Answer:
RHS = (x + y)(x² – xy + y²) = x(x² – xy + y²) + y(x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
– x³ + y³ = L.H.S
Hence proved.

(ii) x³ – y³ = (x – y) (x² + xy + y²)
Answer:
RHS = (x – y)(x² + xy + y²) = x(x² + xy + y²) – y(x² + xy + y²)
= x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³ = LHS Hence proved.

Question 10.
Factorise each of the following :
(i) 27y³ + 125z³
Answer:
27y³ + 125z³ = (3y)³ + (5z)³ = (3y + 5z)[(3y)² – (3y)(5z) + (5z)²]
= (3y + 5z)(9y² – 15yz + 25z²)

(ii) 64m³ – 343n³ (Hint: See question 9)
Answer:
64m³ – 343n³ = (4m)³ – (7n)³ = (4m – 7n)[(4m)² + (4m)(7n) + (7n)²]
= (4m – 7n)(16m² + 28mn + 49n²)

Question 11.
Factorise : 27x³ + y³ + z³ – 9xyz
Answer:
27x³ + y³ + z³ – 9xyz= (3x)³ + y³ + z³ – 3(3x)(y)(z)
= (3x + y + z)[(3x)² + y² + z² – (3x)y – yz – z(3x)] = (3x +y +z)(9x² + y² +z² – 3xy -yz – 3zx)

Question 12.
Verify that x³ + y³ + z³ – 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 + y² + y² – 2yz + z² + z² – 2zx + x³)
= (x + y + z)(x² + y² + z² – yz – zx – xy)
Hence proved.

Question 13.
If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
Answer:
We have : x + y + z = 0 x + y = -z
Cubing both sides, we get:
(x + y)³ = (- z)³
⇒ x³ + y³ + 3xy(x + y) = – z³
⇒ x³ + y³ – 3xyz = -z
⇒ x³ + y³ + z³ – 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
x³ + y³ + z³ = 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
x³ + 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 = 25a² – 35a + 12 = 25a² – 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 = 35y² + 13y – 12 = 35y² + 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 = 3a² – 12a = 3a(a – 4)
∴ Possible expressions for the dimensions of cuboid are 3, a and (x – 4) units.

(ii) Volume = 12ky² + 8ky – 20k = 4k(3y² + 2y – 5)
= 4k(3y² – 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.