HBSE 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

Haryana State Board HBSE 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 2 Polynomials Exercise 2.5

Question 1.
Use suitable identities to find the following products :
(i) (x + 4)(x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (y² + \(\frac {3}{2}\)) (y² – \(\frac {3}{2}\))
(v) (3 – 2x) (3 + 2x).
Solution:
(i) (x + 4) (x + 10) = x² + (4 + 10)x + 4 × 10 (Using the identity, (x + a) (x +b) = x² + (a + b)x + ab)
= x² + 14x + 40.
Hence,
(x + 4) (x + 10) = x² + 14x + 40.

(ii) (x + 8) (x – 10) = x² + [8 + (-10)]x + 8 × (-10)
(Using the identity, (x + a) (x + b) = x² + (a + b)x + ab)
= x² – 2x – 80
Hence,
(x + 8) (x – 10) = x² – 2x – 80.

(iii) (3x + 4) (3x – 5) = (3x)² + [4 + (-5)] × 3x + 4 × (-5)
(Using the identity, (x + a) (x + b) = x² + (a + b)x + ab]
= 9x² – 3x – 20.
Hence,
(3x + 4) (3x – 5) = 9x² – 3x – 20.

(iv) (y² + \(\frac {3}{2}\)) (y² – \(\frac {3}{2}\))
= (y²)² – (\(\frac {3}{2}\))²
(Using the identity,
(x + y) (x – y) = x² – y²)
= y⁴ – \(\frac {9}{4}\)
Hence,
(y² + \(\frac {3}{2}\)) (y² – \(\frac {3}{2}\)) = y⁴ – \(\frac {9}{4}\)

(v) (3 – 2x) (3 + 2x) = (3)² – (2x)²
(Using the identity,
(x + y)(x – y) = x² – y²]
= 9 – 4x²
Hence, (3 – 2x) (3 + 2x) = 9 – 4x²

Question 2.
Evaluate the following products without multiplying directly :
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96.
Solution:
(i) 103 × 107 = (100 + 3) (100 + 7)
= (100)² + (3 + 7) × 100 + 3 × 7
(Using the identity, (x + a) (x + b) = x² + (a + b)x + ab)
= 10000 + 10 × 100 + 21
= 10000 + 1000+ 21
= 11021.
Hence, 103 × 107 = 11021.

(ii) 95 × 96 = (100 – 5)(100 – 4)
=(100)² + [(-5) + (-4)] × 100 + (-5) × (-4)
(Using the identity, (x + a) (x + b) = x² + (a + b)x + ab)
= 10000 – 9 × 100 + 20
= 10000 – 900 + 20 = 9120.
Hence, 95 × 96 = 9120.

(iii) 104 × 96 = (100 + 4) (100 – 4)
= (100)² – (4)² [Using the identity, (x + y) (x – y) = x² – y²]
= 10000 – 16 = 9984.
Hence, 104 × 96 = 9984.

Question 3.
Factorise the following using appropriate identities :
(i) 9x² + 6xy + y²
(ii) 4y² – 4y + 1
(iii) x² – \(\frac{y^2}{100}\)
Solution:
(i) 9x² + 6xy + y² = (3x)² + 2 × 3x × y + y²
= (3x + y)².
[Using the identity, x² + 2xy + y² = (x + y)²]
= (3x + y) (3x + y).
Hence,
9x² + 6xy + y² = (3x + y) (3x + y).

(ii) 4y² – 4y + 1 = (2y)² – 2 × 2y × 1 + (1)²
= (2y – 1)²
(Using the identity. x² – 2xy + y² = (x – y)²]
= (2y – 1) (2y – 1).
Hence,
4y² – 4y + 1 = (2y – 1) (2y – 1).

(iii) x² – \(\frac{y^2}{100}\) = (x)² – (\(\frac {y}{10}\))²
= (x + \(\frac {y}{10}\)) (x – \(\frac {y}{10}\))
[Using the identity,
x² – y² = (x + y) (x – y)]
Hence x² – \(\frac{y^2}{100}\) = (x + \(\frac {y}{10}\)) (x – \(\frac {y}{10}\))

Question 4.
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
(vi) [\(\frac {1}{4}\)a – \(\frac {1}{2}\)b + 1]²
Solution:
We will use the following identity in each
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx 3
(i) (x + 2y + 4z)² = x² + (2y)² + (4z)² + 2 × x × 2y + 2 × 2y × 4z + 2 × 4z × x
= x² + 4y² + 16z² + 4xy + 16yz + 8zx.
Hence,
(x + 2y + 4z)² = x² + 4y² + 16z² + 4xy + 16yz + 8zx.

(ii) (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.
Hence, (2x – y + z)² = 4x² + y² + z² – 4xy – 2yz + 4zx.

(iii) (-2x + 3y + 2z)² = [(-2x) + 3y + 2z]²
= (-2x)² + (3y)² + (2z)² + 2 × (-2x) × 3y + 2 × 3y × 2 + 2 × 2z × (- 2x)
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx.
Hence,(-2x + 3y + 2z)² = 4x² + 9y² + 4z² – 12xy + 12yz – 8zx.

(iv) (3a – 7b – c)² = [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.
Hence,(3a – 7b – c)2 = 9a² + 49b² + c² – 42ab + 14bc – 6ca.

(v) (-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.
Hence,(-2x + 5y – 3z)² = 4x² + 25y² + 9z² – 20xy – 30yz + 12zx.

Question 5.
Factorise :
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz – 8xz.
Solution:
(i) 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
[Using the identity, x² + y² + z² + 2xy + 2yz + 2zx .
= (x + y + z)²
= (2x + 3y – 4z)²
= (2x + 3y – 4z) (2x + 3y – 4z)
Hence, 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
= (2x + 3y – 4z) (2x + 3y – 4z).

(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4 \(\sqrt{2}\)yz – 8xz
= (- \(\sqrt{2}\)x)² + (y)² + (2\(\sqrt{2}\)z)² + 2 × (-\(\sqrt{2}\)x) × (y) + 2 × (y) × (2\(\sqrt{2}\)z) + 2 × (2\(\sqrt{2}\)z) × (-\(\sqrt{2}\)x)
= [(-\(\sqrt{2}\)x) + (y) + (2\(\sqrt{2}\)z)]²
[Using the identity, x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²]
= (- \(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)²
Hence, 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz – 8xz
= (- \(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z) (- \(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)

Question 6.
Write the following cubes in expanded form:
(i) (2x + 1)³
(ii) (2a – 3b)³
(iii) [\(\frac {3}{2}\)x + 1]²
(iv) [x – \(\frac {2}{3}\)y]³
Solution:
(i) (2x + 1)³ = (2x)³ + 1³ + 3 × 2x × 1(2x + 1)
(Using the identity, (x + y)³ = x³ + y³ + 3xy(x + y)]
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 1 + 12x² + 6x.
Hence, (2x + 1)³ = 8x³ + 12x² + 6x + 1.

(ii) (2a – 3b)³ = (2a)³ – (3b)³ – 3 × 2a × (3b) (2a – 3b)
[Using the identity, (x – y)³ = x³ – y³ – 3xy(x – y)]
= 8a³ – 27b³ – 18ab(2a – 3b)
= 8a³ – 27b³ – 36a²b + 54ab².
Hence,
(2a – 3b)³ = 8a³ – 27b³ – 36a²b + 54ab².

Question 7.
Evaluate the following using suitable identities :
(i) (99)³
(ii) (102)³
(ii) (998)³
Solution:
(i) (99)³ = (100 – 1)³
= (100)³ – (1)³ – 3 × 100 × 1(100 – 1)
[Using the identity, (x – y)³ = x³ – y³ – 3xy(x – y)]
= 1000000 – 1 – 300(99)
= 1000000 – 1 – 29700
= 970299.
Hence, (99)³ = 970299.

(ii) (102)³ = (100 + 2)³
= (100)³ + (2)³ + 3 × 100 × 2(100 + 2)
[Using the identity, (x + y)³ = x³ + y³ + 3xy(x + y)]
= 1000000 + 8 + 600(102)
= 1000000 + 8 + 61200
= 1061208.
Hence, (102)³ = 1061208. Ans.
(998)³ = (1000 – 2)³
= (1000)³ – (2)³ – 3 × 1000 × 2(1000 – 2)
(Using the identity,
(x – y)³ = x³ – y³ – 3xy(x – y)]
= 1000000000 – 8 – 6000 × 998
= 1000000000 – 8 – 5988000 = 994011992.
Hence, (998)³ = 994011992.

Question 8.
Factorise each of the following:
(i) 8a³ + b³ + 12a²b + 6ab²
(ii) 8a³ – b³ – 12a²b + 6ab²
(iii) 27 – 125a³ – 135a + 225a²
(iv) 64a³ – 27b³ – 144a²b + 108ab²
(v) 27p³ – \(\frac{1}{216}-\frac{9}{2}\)p² + \(\frac {1}{4}\)p
Solution:
(i) 8a³ + b³ + 120a²b + 6ab²
= (2a)³ + (b)³ + 6ab(2a + b)
= (2a)³ + (b)³ + 3 × 2a × b(2a + b)
= (2a + b)³
[Using the identity, x³ + y³ + 3xy(x + y) = (x + y)³]
= (2a + b) (2a + b) (2a + b).
Hence, 8a³ + b³ + 12a²b + 6ab² = (2a + b) (2a +b) (2a + b).

(ii) 8a³ – b³ – 12a²b + 6ab² = (2a)³ – (b)³ – 6ab(2a – b)
= (2a)³ – (b)³ – 3 × 2a × b(2a – b) = (2a -b)
[Using the identity,
x³ – y³ – 3xy(x – y) = (x – y)³]
= (2a – b) (2a – b) (2a – b).
Hence, 8a³ – b³ – 12a²b + 6ab² = (2a – b) (2a – b).

(iii) 27 – 125a³ – 135a + 225a² = (3)³ – (5a)³ – 45a(3 – 5a)
= (3)³ – (5a)³ – 3 × 3 × 5a(3 – 5a)
= (3 – 5a)³
[Using the identity,
x³ – y³ – 3xy(x – y) = (x – y)³] = (3 – 5a) (3 – 5a) (3 – 5a). Hence, 27 – 125a³ – 135a + 225a² = (3 – 5a) (3 – 5a) (3 – 5a).

(iv) 64a³ – 27b³ – 144a²b + 108ab²
= (4a)³ – (3b)³ – 36ab(4a – 3b)
= (4a)³ – (3b)³ – 3 × 4a × 3b(4a – 3b)
= (4a – 3b)³
[Using the identity,
x³ – y³ – 3xy(x – y) = (x – y)³)
= (4a – 3b) (4a – 3b) (4a – 3b).
Hence, 64a³ – 27b³ – 144a²b + 108ab²
= (4a – 3b) (4a – 3b) (4a – 3b).

Question 9.
Verify:
(i) x³ + y³ = (x + y) (x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²).
Solution:
(i) We have,
x³ + y³ = (x + y) (x² – xy + y²)
R.H.S. = (x + y) (x² – xy + y²)
= x(x² – xy + y²) + y(x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³ = LH.S.
Hence, L.H.S. = R.H.S. Verified

(ii) We have x³ – y³ = (x – y) (x² + xy + y²)
R.H.S = (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, L.H.S. = R.H.S. Verified

Question 10.
Factorise each of the following:
(i) 27y³ + 125z³
(ii) 64m³ – 343n³ (Hint : See question 9).
Solution:
(i) 27y³ + 125z³ = (3y)³ + (5z)³
= (3y + 5z) [(3y)² – 3y × 5z + (5z)²]
[Using the identity,
x³ + y³ = (x + y) (x² – xy + y2)]
= (3y + 5z) (9y² – 15yz² + 25z²).
Hence, 27y³ + 125z³ = (3y + 5z) (9y² – 15yz + 25z²).

(ii) 64m³ – 343n³ = (4m)³ – (7n)³
= (4m – 7n) [(4m)² + 4m × 7n + (7n)²]
[Using the identity,
x³ – y³ = (x – y) (x² + xy + y²)]
= (4m – 7n) (16m² + 28mn + 49n²).
Hence, 64m³ – 343n³ = (4m – 7n) (16m² + 28mn + 49n²).

Question 11.
Factorise : 27x³ + y³ + z³ – 9xyz.
Solution :
27x³ + y³ + z³ – 9xyz = (3x)³ + (y)³
+ (z)³ – 3 × 3x × y × z
= (3x + y + z) [(3x)² + y² + z² – 3x × y – y × z – z × 3x)
[Using the identity,
x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)]
= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3zx).
Hence, 27x³ + y³ + z³ – 9xyz = (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).
Solution:
We have, x³ + y³ + z² – 3xyz = \(\frac {1}{2}\)(x + y + z) [(x – y)² + (y – z)² + (z – x)²).
R.H.S. = \(\frac {1}{2}\)(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 – 2xz + x²)
= \(\frac {1}{2}\)(x + y + z) (2x² + 2y² + 2z² – 2xy – 2yz – 2xz)
= \(\frac {1}{2}\) × 2(x + y + z) (x² + y² + z² – xy – yz – xz)
= x³ + y³ + z³ – 3xyz.
[Using the identity, (x + y + z)(x² + y² + z² – xy – yz – xz]
= x³ + y³ + z³ – 3xyz]
= L.H.S.
Hence, L.H.S. = R.H.S. Verified

Question 13.
If x + y + z = 0, show that : x³ + y³ + z³ = 3xyz.
Solution:
We have,
x + y + z = 0
⇒ x + y = – z …(i)
Cubing both sides of equation (i), we get
(x + y)³ = (-z)³
⇒ x³ + y³ + 3xy(x + y) = -z³
⇒ x³ + y³ + z³ + 3xy × (-z) = 0, [∵ x + y = -z]
⇒ x³ + y³ + z³ – 3xyz = 0
⇒ x³ + y³ + z³ = 3xyz. Proved

Question 14.
Without actually calculating the cubes, find the value of each of the following:
(i) (- 12)³ + (7)³ + (5)³,
(ii) (28)³ + (-15)³ + (-13)³.
Solution:
(i) We know that, if
x + y + z = 0, then
x³ + y³ + z³ = 3xyz
Since,
(-12) + 7 + 5 = -12 + 12 = 0
Therefore, (- 12)³ + (7)³ + (5)³ = 3 × (-12) × 7 × 5
= – 1260
Hence,
(- 12)³ + (7)³ + (5)³ = -1260.

(ii) We know that, If x + y + z = 0, then
x³ + y³ + z³ = 3xyz
Since, 28 + (-15) + (-13) = 28 – 28 = 0
Therefore, (28)³ + (-15)³ + (-13)³
= 3 × 28 × (-15) × (-13)
= 16380.
Hence,
(28)³ + (-15)³ + (-13)³ = 16380.

Question 15.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a² – 35a + 12
(ii) Area : 35y² + 13y – 12
Solution:
(i) We have,
Area of rectangle = 25a² – 35a + 12
[∵ 20 × 15 = 300
20 + 15 = 35]
⇒ length × breadth = 25a² – (20 + 15)a + 12
⇒ length × breadth = 25a² – 20a – 15a + 12
⇒ length × breadth = (25a² – 20a) – (15a – 12)
⇒ length × breadth = 5a(5a – 4) – 3(5a – 4)
⇒ length × breadth = (5a – 4) (5a – 3)
⇒ length × breadth = (5a – 3) (5a – 4)
Hence, One possible answer is length= (5a – 3) and breadth = (5a – 4).

(ii) We have,
Area of rectangle = 35y² + 13y – 12
[∵ 28 × (-15) = – 420
28 – 15 = 13]
⇒ length × breadth = 35y² + (28 -15)y – 12,
⇒ length × breadth = 35y² + 28y – 15y – 12
⇒ length × breadth = (35y² + 28y) – (15y + 12)
⇒ length × breadth = 7y(5y + 4) – 3(5y + 4)
⇒ length × breadth = (5y + 4) (7y – 3)
⇒ length × breadth = (7y – 3) (5y + 4)
Hence, one possible answer is:
length = (7y – 3) and breadth = (5y + 4).

Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume: 3x² – 12x
(ii) Volume: 12ky² + 8ky – 20k
Solution:
(i) We have,
Volume of cuboid = 3x² – 12x
⇒ length × breadth × height = 3x (x – 4)
⇒ length × breadth × height = 3 × x × (x – 4)
Hence, possible dimensions of cuboid are 3, x and (x – 4).

(ii) We have,
Volume of cuboid = 12ky² + 8ky – 20k
⇒ length × breadth × height = 4k[3y² + 2y – 5] [∵ 5 × (-3) = – 15, 5 – 3 = 2]
⇒ length × breadth × height = 4k[3y² + (5 – 3)y – 5)]
⇒ length × breadth × height = 4k[3y² + 5y – 3y – 5]
⇒ length × breadth × height = 4k[(3y² + 5y) – (3y + 5)]
⇒ length × breadth × height = 4k[y(3y + 5) – 1(3y + 5)]
⇒ length × breadth × height = 4k × (3y + 5) (y – 1)
Hence, possible demensions of cuboid are 4k, (3y + 5) and (y-1).