# NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths 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) We have given, (x + 4)(x + 10)
We know that
(x + a) (x + b) = x2 + (a + b)x + ab
Therefore,
(x + 4) (x + 10)
= x2 + (4 + 10)x + 4 × 10
= x2 + 14x + 40

(ii) We have given (x + 8) (x – 10)
or, (x + 8) (x + (-10))
We know that,
(x + a) (x + b) = x2 + (a + b)x + ab
(x + 8) (x + (-10))
= x2 + (8 + (-10))x + 8 × (-10)
= x2 – 2x – 80 (iii) We have given, (3x + 4)(3x – 5)
or, (3x + 4) (3x + (- 5))
We know that
(x + a) (x + b) = x2 + (a + b)x + ab
(3x + 4)(3x – 5)
= (3x)2 + (4 + (-5))3x + 4 × (-5)
= 9x2 – 3x – 20

(iv) We have given (y2 + $$\frac{3}{2}$$) (y2 – $$\frac{3}{2}$$)
We know that (x + y)(x – y) = x2 – y2
$$\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}$$
= $$y^{4}-\frac{9}{4}$$

(v) We have given (3 – 2x)(3 + 2x)
We know that (x – y)(x + y) = x2 – y2
(3 – 2x)(3 + 2x)
= (3)2 – (2x)2
= 9 – 4x2 Question 2.
Evaluate the following products without multiplying directly.
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
Solution:
(i) We have given, 103 × 107
We can also write
(100 + 3) × (100 + 7)
We know that
(x + a) (x + b) = x2 + (a + b)x + ab
(100 + 3) (100 + 7)
= (100)2 + (3 + 7) × 100 + 3 × 7
= 10000 + 1000 + 21
= 11021

(ii) We have given, 95 × 96
We can also write (90 + 5) × (90 + 6)
We know that
(x + a) (x + b) = x2 + (a + b)x + ab
(90 + 5) × (90 + 6)
= (90)2 + (5 + 6) × 90 + 5 × 6
= 8100 + 990 + 30
= 9120 (iii) We have given, 104 × 96
We can also write (100 + 4) × (100 – 4)
We know that (x + y)(x – y) = x2 – y2
(100 + 4)(100 – 4)
= (100)2 – (4)2
= 10000 – 16
= 9984

Question 3.
Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2
(ii) 4y2 – 4y + 1
(iii) $$x^{2}-\frac{y^{2}}{100}$$
Solution:
(i) We have given, 9x2 + 6xy + y2
We can also write (3x)2 + 2 . 3x . y + (y)2
We know that
a2 + 2ab + b2 = (a + b)2
(3x)2 + 2(3x)(y) + (y)2
= (3x + y)2
= (3x + y)(3x + y)

(ii) We have, 4y2 – 4y + 1
We can also write (2y)2 – 2 . 2y . 1 + (1)2
We know that
x2 – 2xy + y2 = (x – y)2
(2y)2 – 2 . 2y . 1 + (1)2
= (2y – 1)2
= (2y – 1)(2y – 1)

(iii) We have given, $$x^{2}-\frac{y^{2}}{100}$$
We can also write, $$(x)^{2}-\left(\frac{y}{100}\right)^{2}$$
We know that
a2 – b2 = (a + b)(a – b)
$$(x)^{2}-\left(\frac{y}{100}\right)^{2}=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)$$ 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
(iv) (-2x + 5y – 3z)2
(v) $$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$$
Solution:
(i) We have given that, (x + 2y + 4z)2
We know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Therefore,
(x + 2y + 4z)2
= (x)2 + (2y)2 + (4z)2 + 2 . x . 2y + 2 . 2y . 4z + 2 . 4z . x
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

(ii) We have given that (2x – y + z)2
We can also write, (2x + (-y) + z)2
Now, we know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Therefore,
(2x + (-y) + z)2
= (2x)2 + (-y)2 + (z)2 + 2 . 2x . (-y) + 2 . (-y) . z + 2 . z . 2x
= 4x2 + y2 + z2 – 4xy – 2yz + 4zx (iii) We have given that, (-2x + 3y + 2z)2
We can also write, [(-2x) + 3y + 2z]2
Now, we know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Therefore,
((-2x) + 3y + 2z)2 = (-2x)2 + (3y)2 + (2z)2 + 2 . (-2x) . 3y + 2 . 3y . 2z + 2 . 2z . (-2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx

(iv) We have given that, (3a – 7b – c)2
We can also write (3a + (-7b) + (-c)2
Now, we know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Therefore,
(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

(v) We have given that, (-2x + 5y – 3z)2
We can also write [(-2x) + 5y + (-3z)]2
Now, we know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Therefore,
[(-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) We have given that, $$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$$
We can also write $$\left[\frac{1}{4} a\left(-\frac{1}{2} b\right)+1\right]^{2}$$
Now, we know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Therefore, Question 5.
Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz
Solution:
(i) We have given
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
We can also write
(2x)2 + (3y)2 + (-4z)2 + 2 . 2x . 3y + 2 . 3y . (-4z) + 2 . (-4z) . 2x
Now we know that
a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2
Therefore,
(2x)2 + (3y)2 + (-4z)2 + 2 . 2x . 3y + 2 . 3y (-4z) + 2 . (-4z). 2x
= (2x + 3y + (-4z))2
= (2x + 3y – 4z)2
= (2x + 3y – 4z) (2x + 3y – 4z)

(ii) We have given that
2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz
We can also write
(-√2x)2 + (y)2 + (2√2z)2 + 2 . (-√2x) . y + 2 . y . (2√2z) + 2 . (2√2 z) . (-√2x)
Now, we know that
a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2
Therefore,
(-√2x)2 + (y)2 + (2√2z)2 + 2 . (-√2x) . y + 2 . y . (2√2z) + 2 . (2√2z) . (-√2x)
= ((-√2x) + y + 2√2z)2
= (-√2x + y + 2√2z) (-√2x + y + 2√2z) Question 6.
Write the following cubes in expanded front:
(i) (2x + 1)3
(ii) (2a – 3b)3
(iii) $$\left[\frac{3}{2} x+1\right]^{3}$$
(iv) $$\left[x-\frac{2}{3} y\right]$$
Solution:
(i) We have given that, (2x + 1)3
We know that
(a + b)3 = a3 + b3 + 3ab(a + b)
Therefore,
(2x + 1)3 = (2x)3 + (1)3 + 3 . (2x)(1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 1 + 12x3 + 6x
= 8x3 + 12x3 + 6x + 1

(ii) We have given, (2a – 3b)3
We know that (a – b)3 = a3 – b3 – 3ab(a – b)
(2a – 3b)3
= 8a3 – 27b3 – 18ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2

(iii) We have given that, $$\left[\frac{3}{2} x+1\right]^{3}$$
We know that
(a + b)3 = a3 + b3 + 3ab(a + b)
Therefore, (iv) We have given that, $$\left[x-\frac{2}{3} y\right]$$
We know that
(a – b)3 = a3 – b3 – 3ab(a – b)
Therefore, Question 7.
Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) We have given, (99)3
We can also write, (100 – 1)3
We know that (a – b)3 = a3 – b3 – 3ab(a – b)
(100 – 1)3 = (100)3 – (1)3 – 3 . 100 . 1 (100 – 1)
= 10,00,000 – 1 – 30000 + 300
= 9,70,299

(ii) We have given that, (102)3
We can also write (100 + 2)3
Now, we know that
(a + b)3 = a3 + b3 + 3ab(a + b)
(100 + 2)3 = (100)3 + (2)3+ 3 . 100 . 2 (100 + 2)
= 10,00,000 + 8 + 600(100 + 2)
= 10,00,000 + 8 + 60000 + 1200
= 10,61,208

(iii) We have given that (998)3
We can also write (1000 – 2)3
Now, we know that (a – b)3 = a3 – b3 – 3ab(a – b)
(1000 – 2)3 = (1000)3 – (2)3 – 3 . 1000 . 2(1000 – 2)
= 1,00,00,00,000 – 8 – 6000 (100 – 2)
= 99,40,11,992 Question 8.
Factarise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a3 – b3 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a3
(iv) 64a3 – 27b3 -144a2b + 108ab2
(v) $$27 \mathrm{P}^{3}-\frac{1}{216}-\frac{9}{2} \mathrm{P}^{2}+\frac{1}{4} \mathrm{P}$$
Solution:
(i) We have given that
8a3 + b3 + 12a2b + 6ab2
We can also write
(2a)3 + (b)3 + 6ab(2a + b)
or, (2a)3 + b3 + 3 . 2a . b(2a + b)
We know that
a3 + b3 + 3ab(a + b) = (a + b)3
(2a)3 + (b)3 + 3 . 2a . b (2a + b) = (2a + b)3

(ii) We have given that
8a3 – b3– 12a2b + 6ab2
We can also write
(2a)3 – (b)3 – 6ab(2a – b)
or, (2a)3 – (b)3 – 3 . 2a . b(2a – b)
We know that
a3 – b3 – 3ab(a – b) = (a – b)3
(2a)3 – (b)3 – 3 . 2a . b(2a – b) = (2a – b)3 (iii) We have given that
27 – 125a3 – 135a + 225a2
We can also write (3)3 – (5a)3 – 45a(3 – 5a)
or, (3)3 – (5a)3 – 3 . 3 . 5a(3 – 5a)
Now, we know that,
a3 – b3 – 3ab(a – b) = (a – b)3
(3)3 – (5a)3 – 3 . 3 . 5a(3 – 5a)
= (3 – 5a)3
= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) We have given that
64a3 – 27b3 – 144a2b + 108ab2
We can also write
(4a)3 – (3b)3 – 36ab(4a – 3b)
or, (4a)3 – (3b)3 – 3 . 4a . 3b(4a – 3b)
Now, we know that
a3 – b3 – 3ab(a – b) = (a – b)3
(4a)3 – (3b)3 – 3 . 4a . 3b(4a – 3b)
= (4a – 3b)3

(v) We have given that
$$27 \mathrm{P}^{3}-\frac{1}{216}-\frac{9}{2} \mathrm{P}^{2}+\frac{1}{4} \mathbf{P}$$
We can also write,
$$(3 P)^{3}-\left(\frac{1}{6}\right)^{3}-3.3 P \cdot \frac{1}{6}\left(3 P-\frac{1}{6}\right)$$
We know that
a3 – b3 – 3ab(a – b) = (a – b)3
$$(3 P)^{3}-\left(\frac{1}{6}\right)^{3}-3.3 P \cdot \frac{1}{6}\left(3 P-\frac{1}{6}\right)=\left(3 P-\frac{1}{6}\right)^{3}$$ Question 9.
Verify
(i) x3 + y3 = (x + y) (x2 – xy + y2)
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
(i) R.H.S.
(x + y) (x2 – xy + y2)
By actual multiplication we have
x3 – x2y + xy2 + x2y – xy2 + y3
= x3 + y3
= L.H.S

(ii) R.H.S.
(x – y)(x2 + xy + y2)
By actual multiplication we have
x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3
= L.H.S.

Question 10.
Factorise each of the following:
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
Solution:
(i) We have given, 27y3 + 125z3
We can also write, (3y)3 + (5z)3
We know that
a3 + b3 = (a + b)(a2 – ab + b2)
Therefore,
(3y)3 + (5z)3 = (3y + 5z) ((3y)2 – (3y).(5z) + (5z)2)
= (3y + 5z) (9y2 – 15yz + 25z2)

(ii) We have given, 64m3 – 343n3
We can also write (4m)3 – (7n)3
We know that,
a3 – b3 = (a – b)(a2 + ab + b2)
Therefore,
(4m)3 – (7n)3 = (4m – 7n)((4m)2 + (4m)(7n) + (7n)2)
= (4m – 7n)(16m2 + 28mn + 49n2) Question 11.
Factorise: 27x3 + y3 + z3 – 9xyz
Solution:
We have given,
27x3 + y3 + z3 – 9xyz
We can also write,
(3x)3 + (y)3 + (z)3 – 3 . (3x) . (y) . (z)
We know that
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
Therefore,
(3x)3 + (y)3 + (z)3 – 3 . (3x) (y) (z)
= (3x + y + z) [(3x)2 + (y)2 + (z)2 – (3x)(y) -(y)(z) – (z)(3x)]
= (3x + y + z) (9x2 + y2 + z2 – 3xy – yz – 3zx)

Question 12.
Verify that x3 + y3 + z3 – 3xyz = 1/2 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
Solution:
We have given L.H.S. is x3 + y3 + z3 – 3xyz
We know that
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
Multiply and divide 2 we get
= (x + y + z)2 × $$\frac{1}{2}$$ (x2 + y2 + z2 – xy – yz – zx)
= $$\frac{1}{2}$$ (x + y + z)(2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx)
= $$\frac{1}{2}$$ (x + y + z) (x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx)
= $$\frac{1}{2}$$ (x + y + z)((x – y)2 + (y – z)2 + (z – x)2)
Hence verified. Question 13.
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Solution:
We know that,
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Here, we have given x + y + z = 0
x3 + y3 + z3 – 3xyz = 0 × (x2 + y2 + z2 – xy – yz – zx)
or, x3 + y3 + z3 – 3xyz = 0
x3 + y3 + z3 = 3xyz

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 have given that (-12)3 + (7)3 + (5)3
First we check the value of x + y + z
We have -12 + 7 + 5 = 0
We know that if x + y + z = 0 then x3 + y3 + z3 = 3xyz
(-12)3 + (7)3 + (5)3 = 3 × (-12) × 7 × 5 = -1260

(ii) We have given that
(28)3 + (-15)3 + (-13)3
First we check the value of x + y + z
We have 28 + (-15) + (-13) = 0
We know that if x + y + z = 0 then,
x3 + y3 + z3 = 3xyz
(28)3 + (-15)3 + (-13)3
= 3 × 28 × (-15) × (-13)
= 16,380 Question 15.
Give possible expressions of 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 know that
Area of rectangle = length × breadth …. (i)
But, we have given that
Area of rectangle = 25a2 – 35a + 12 …. (ii)
from equation (i) and (ii)
l × b = 25a2 – 35a + 12
= 25a2 – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
l × b = (5a – 4)(5a – 3)
length (l) = (5a – 4) and breadth (b) = (5a – 3)

(ii) We know that
Area of rectangle = length × breadth ……. (i)
But we have given that
Area of rectangle = 35y2 + 13y – 12 ……. (ii)
From equation (i) and (ii)
length × breadth = 35y2 + 13y – 12
= 35y2 + 28y – 15y – 12
= 7y(5y + 4) – 3(5y + 4)
or, length × breadth = (5y + 4) (7y – 3)
Therefore,
length = (5y + 4) Question 16.
What are the possible expression for the dimensions of the cuboids whose volumes are given below:
(i) volume = 3x2 – 12x
(ii) volume = 12ky2 + 8ky – 20k
Solution:
(i) We know that
Volume of cuboid = length × breadth × height ……. (i)
But we have given that
Volume of cuboid = 3x2 – 12x …… (ii)
From equation (i) and (ii)
length × breadth × height = 3x2 – 12x = 3x(x – 4)
or, length × breadth × height = 3 × x × (x – 4)
length of cuboid = 3
height of cuboid = (x – 4)

(ii) We know that,
Volume of cuboid = length × breadth × height …….. (i)
But, we have given that Volume of cuboid = 12ky2 + 8ky – 20k ……. (ii)
From equation (i) and (ii)