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

(v) (3 – 2x) (3 + 2x)

Solution:

(i) We have given, (x + 4)(x + 10)

We know that

(x + a) (x + b) = x^{2} + (a + b)x + ab

Therefore,

(x + 4) (x + 10)

= x^{2} + (4 + 10)x + 4 × 10

= x^{2} + 14x + 40

(ii) We have given (x + 8) (x – 10)

or, (x + 8) (x + (-10))

We know that,

(x + a) (x + b) = x^{2} + (a + b)x + ab

(x + 8) (x + (-10))

= x^{2} + (8 + (-10))x + 8 × (-10)

= x^{2} – 2x – 80

(iii) We have given, (3x + 4)(3x – 5)

or, (3x + 4) (3x + (- 5))

We know that

(x + a) (x + b) = x^{2} + (a + b)x + ab

(3x + 4)(3x – 5)

= (3x)^{2} + (4 + (-5))3x + 4 × (-5)

= 9x^{2} – 3x – 20

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

We know that (x + y)(x – y) = x^{2} – y^{2}

\(\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\)

= \(y^{4}-\frac{9}{4}\)

(v) We have given (3 – 2x)(3 + 2x)

We know that (x – y)(x + y) = x^{2} – y^{2}

(3 – 2x)(3 + 2x)

= (3)^{2} – (2x)^{2}

= 9 – 4x^{2}

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

(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}

(ii) 4y^{2} – 4y + 1

(iii) \(x^{2}-\frac{y^{2}}{100}\)

Solution:

(i) We have given, 9x^{2} + 6xy + y^{2}

We can also write (3x)^{2} + 2 . 3x . y + (y)^{2}

We know that

a^{2} + 2ab + b^{2} = (a + b)^{2}

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

= (3x + y)^{2}

= (3x + y)(3x + y)

(ii) We have, 4y^{2} – 4y + 1

We can also write (2y)^{2} – 2 . 2y . 1 + (1)^{2}

We know that

x^{2} – 2xy + y^{2} = (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

a^{2} – b^{2} = (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} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Therefore,

(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) We have given that (2x – y + z)^{2}

We can also write, (2x + (-y) + z)^{2}

Now, we know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Therefore,

(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) We have given that, (-2x + 3y + 2z)^{2}

We can also write, [(-2x) + 3y + 2z]^{2}

Now, we know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Therefore,

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

= 4x^{2} + 9y^{2} + 4z^{2} – 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} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Therefore,

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

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

= 9a^{2} + 49b^{2} + c^{2} – 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} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Therefore,

[(-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) 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} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Therefore,

Question 5.

Factorise:

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

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

Solution:

(i) We have given

4x^{2} + 9y^{2} + 16z^{2} + 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

a^{2} + b^{2} + c^{2} + 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

2x^{2} + y^{2} + 8z^{2} – 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

a^{2} + b^{2} + c^{2} + 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} = a^{3} + b^{3} + 3ab(a + b)

Therefore,

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

= 8x^{3} + 1 + 6x(2x + 1)

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

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

(ii) We have given, (2a – 3b)^{3}

We know that (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

(2a – 3b)^{3}

= 8a^{3} – 27b^{3} – 18ab(2a – 3b)

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

(iii) We have given that, \(\left[\frac{3}{2} x+1\right]^{3}\)

We know that

(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

Therefore,

(iv) We have given that, \(\left[x-\frac{2}{3} y\right]\)

We know that

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

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

(iii) 27 – 125a^{3} – 135a + 225a^{3}

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

(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

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

We can also write

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

or, (2a)^{3} + b^{3} + 3 . 2a . b(2a + b)

We know that

a^{3} + b^{3} + 3ab(a + b) = (a + b)^{3}

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

(ii) We have given that

8a^{3} – b^{3}– 12a^{2}b + 6ab^{2}

We can also write

(2a)^{3} – (b)^{3} – 6ab(2a – b)

or, (2a)^{3} – (b)^{3} – 3 . 2a . b(2a – b)

We know that

a^{3} – b^{3} – 3ab(a – b) = (a – b)^{3}

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

(iii) We have given that

27 – 125a^{3} – 135a + 225a^{2}

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,

a^{3} – b^{3} – 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

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

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

a^{3} – b^{3} – 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

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

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

Solution:

(i) R.H.S.

(x + y) (x^{2} – xy + y^{2})

By actual multiplication we have

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

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

= L.H.S

(ii) R.H.S.

(x – y)(x^{2} + xy + y^{2})

By actual multiplication we have

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

= x^{3} – y^{3}

= L.H.S.

Question 10.

Factorise each of the following:

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

(ii) 64m^{3} – 343n^{3}

Solution:

(i) We have given, 27y^{3} + 125z^{3}

We can also write, (3y)^{3} + (5z)^{3}

We know that

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

Therefore,

(3y)^{3} + (5z)^{3} = (3y + 5z) ((3y)^{2} – (3y).(5z) + (5z)^{2})

= (3y + 5z) (9y^{2} – 15yz + 25z^{2})

(ii) We have given, 64m^{3} – 343n^{3}

We can also write (4m)^{3} – (7n)^{3}

We know that,

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

Therefore,

(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

Solution:

We have given,

27x^{3} + y^{3} + z^{3} – 9xyz

We can also write,

(3x)^{3} + (y)^{3} + (z)^{3} – 3 . (3x) . (y) . (z)

We know that

a^{3} + b^{3} + c^{3} – 3abc = (a + b + c) (a^{2} + b^{2} + c^{2} – 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) (9x^{2} + y^{2} + z^{2} – 3xy – yz – 3zx)

Question 12.

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

Solution:

We have given L.H.S. is x^{3} + y^{3} + z^{3} – 3xyz

We know that

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

Multiply and divide 2 we get

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

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

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

Solution:

We know that,

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

Here, we have given x + y + z = 0

x^{3} + y^{3} + z^{3} – 3xyz = 0 × (x^{2} + y^{2} + z^{2} – xy – yz – zx)

or, x^{3} + y^{3} + z^{3} – 3xyz = 0

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

x^{3} + y^{3} + z^{3} = 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: 25a^{2} – 35a + 12

(ii) Area: 35y^{2} + 13y – 12

Solution:

(i) We know that

Area of rectangle = length × breadth …. (i)

But, we have given that

Area of rectangle = 25a^{2} – 35a + 12 …. (ii)

from equation (i) and (ii)

l × b = 25a^{2} – 35a + 12

= 25a^{2} – 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 = 35y^{2} + 13y – 12 ……. (ii)

From equation (i) and (ii)

length × breadth = 35y^{2} + 13y – 12

= 35y^{2} + 28y – 15y – 12

= 7y(5y + 4) – 3(5y + 4)

or, length × breadth = (5y + 4) (7y – 3)

Therefore,

length = (5y + 4)

breadth = (7y – 3)

Question 16.

What are the possible expression for the dimensions of the cuboids whose volumes are given below:

(i) volume = 3x^{2} – 12x

(ii) volume = 12ky^{2} + 8ky – 20k

Solution:

(i) We know that

Volume of cuboid = length × breadth × height ……. (i)

But we have given that

Volume of cuboid = 3x^{2} – 12x …… (ii)

From equation (i) and (ii)

length × breadth × height = 3x^{2} – 12x = 3x(x – 4)

or, length × breadth × height = 3 × x × (x – 4)

length of cuboid = 3

breadth of cuboid = x

height of cuboid = (x – 4)

(ii) We know that,

Volume of cuboid = length × breadth × height …….. (i)

But, we have given that Volume of cuboid = 12ky^{2} + 8ky – 20k ……. (ii)

From equation (i) and (ii)

length × breadth × height

= 12ky^{2} + 8ky – 20k

= 4k (3y^{2} + 2y – 5)

= 4k (3y^{2} + 5y – 3y – 5)

= 4k[y(3y + 5) – 1(3y + 5)]

or, length × breadth × height = 4k(3y + 5)(y – 1)

length of cuboid = 4k

breadth of cuboid = (3y + 5)

and height of cuboid = (y – 1)