# Class 9 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.5 | Set 1

### Question 1. Use suitable identities to find the following products:

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

Solution:

Using formula, (x + a) (x + b) = x2 + (a + b)x + ab

[So, a = 4 and b = 10]

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

= x2 + 14x + 40

(ii) (x + 8) (x – 10)

Solution:

Using formula, (x + a) (x + b) = x2 + (a + b)x + ab

[So, a = 8 and b = −10]

(x + 8) (x – 10) = x2 + (8 + (-10) )x + (8 × (-10))

= x2 + (8 – 10) x – 80

= x2 − 2x – 80

(iii) (3x + 4) (3x – 5)

Solution:

Using formula, (y + a) (y + b) = y2 + (a + b)y + ab

[So, y = 3x, a = 4 and b = −5]

(3x + 4) (3x − 5) = (3x)2 + [4 + (-5)]3x + 4 × (-5)

= 9x2 + 3x (4 – 5) – 20

= 9x2 – 3x – 20

(iv) (y2 ) (y2 – )

Solution:

Using formula, (a + b) (a – b) = a2 – b2

[So, a = y2 and b = ]

(y 2 ) (y2 – ) = (y2)2 – ( )^2

= y 4 – ### Question 2. Evaluate the following products without multiplying directly:

(i) 103 × 107

Solution:

103 × 107 = (100 + 3) × (100 + 7)

Using formula, (x + a) (x + b) = x2 + (a + b)x + ab

Then,

x = 100

a = 3

b = 7

So, 103 × 107 = (100 + 3) × (100 + 7)

= (100)2 + (3 + 7)100 + (3 × 7)

= 10000 + 1000 + 21

= 11021

(ii) 95 × 96

Solution:

95 × 96 = (100 – 5) × (100 – 4)

Using formula, (x – a) (x – b) = x2 – (a + b)x + ab

Then, According to the identity

x = 100

a = 5

b = 4

So, 95 × 96 = (100 – 5) × (100 – 4)

= (100)2 – 100 (5+4) + (5 × 4)

= 10000 – 900 + 20

= 9120

(iii) 104 × 96

Solution:

104 × 96 = (100 + 4) × (100 – 4)

Using formula, (a + b) (a – b) = a2 – b2

Then,

a = 100

b = 4

So, 104 × 96 = (100 + 4) × (100 – 4)

= (100)2 – (4)2

= 10000 – 16

= 9984

### Question 3. Factorize the following using appropriate identities:

(i) 9x2 + 6xy + y2

Solution:

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

Using formula, a2 + 2ab + b2 = (a + b)2

Then,

a = 3x

b = y

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

= (3x + y)2

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

(ii) 4y2 − 4y + 1

Solution:

4y2 − 4y + 1 = (2y)2 – (2 × 2y × 1) + 1

Using formula, a2 – 2ab + b2 = (a – b)2

Then,

a = 2y

b = 1

= (2y – 1)2

= (2y – 1) (2y – 1)

(iii)  x2 – Solution:

x2 – = x2 – Using formula, a2 – b2 = (a – b) (a + b)

Then,

a = x

b = = (x – ) (x + )

### Question 4. Expand each of the following, using suitable identities:

(i) (x + 2y + 4z)2

Solution:

Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Then,

x = x

y = 2y

z = 4z

(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + (2 × x × 2y) + (2 × 2y × 4z) + (2 × 4z × x)

= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(ii) (2x − y + z)2

Solution:

Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Then,

x = 2x

y = −y

z = z

(2x − y + z)2 = (2x)2 + (−y)2 + z2 + (2 × 2x × −y) + (2 × −y × z) + (2 × z × 2x)

= 4x2 + y2 + z2 – 4xy – 2yz + 4xz

(iii) (−2x + 3y + 2z)2

Solution:

Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Then,

x = −2x

y = 3y

z = 2z

(−2x + 3y + 2z)2 = (−2x)2 + (3y)2 + (2z)2 + (2 ×−2x × 3y) + (2 ×3y × 2z) + (2 ×2z × −2x)

= 4x2 + 9y2 + 4z2 – 12xy + 12yz– 8xz

(iv) (3a – 7b – c)2

Solution:

Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Then,

x = 3a

y = – 7b

z = – c

(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) (–2x + 5y – 3z)2

Solution:

Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Then,

x = –2x

y = 5y

z = – 3z

(–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) ( a – b + 1)2

Solution:

Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Then,

x = a

y = b

z = 1

( a – ( )b + 1)2 = [ a]2 + [ b]2 + 12 + [2 x }a x b] + [2 x b x 1] + [2 x 1 x a] a2 b2 + 1 – ab – b + a

### Question 5. Factorize:

(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz

Solution:

Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Then, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2

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

= (2x + 3y – 4z) (2x + 3y – 4z)

(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz

Solution:

Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Then, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2

2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz

= (-√2x)2 + (y)2 + (2√2z)2 + (2 × -√2x × y) + (2 × y × 2√2z) + (2 × 2√2 × −√2x)

= (−√2x + y + 2√2z)2

= (−√2x + y + 2√2z) (−√2x + y + 2√2z)

### Question 6. Write the following cubes in expanded form:

(i) (2x + 1)3

Solution:

Using formula,(x + y)3 = x3 + y3 + 3xy(x + y)

(2x + 1)3= (2x)3 + 13 + (3 × 2x ×1) (2x + 1)

= 8x3 + 1 + 6x(2x + 1)

= 8x3 + 12x2 + 6x + 1

(ii) (2a − 3b)3

Solution:

Using formula, (x – y)3 = x3 – y3 – 3xy(x – y)

(2a − 3b)3 = (2a)3 − (3b)3 – (3 × 2a × 3b) (2a – 3b)

= 8a3 – 27b3 – 18ab(2a – 3b)

= 8a3 – 27b3 – 36a2b + 54ab2

(iii) ( x + 1)3

Solution:

Using formula, (x + y)3 = x3 + y3 + 3xy(x + y)

( x+ 1)3 = ( x)3 + 13 + (3 × x × 1) ( x + 1) x3 + 1 + x2 x (iv) (x − y)3

Solution:

Using formula, (x – y)3 = x3 – y3 – 3xy(x – y)

(x − y)3 = x3 − [ y]3 – 3(x) y[x − y]

= x3 y3 – 2x2y + xy2

### Question 7. Evaluate the following using suitable identities:

(i) (99)3

Solution:

99 = 100 – 1

Using formula, (x – y)3 = x3 – y3 – 3xy(x – y)

(99)3 = (100 – 1)3

= (100)3 – 13 – (3 × 100 × 1) (100 – 1)

= 1000000 – 1 – 300(100 – 1)

= 1000000 – 1 – 30000 + 300

= 970299

(ii) (102)3

Solution:

102 = 100 + 2

Using formula, (x + y)3 = x3 + y3 + 3xy(x + y)

(100 + 2)3 = (100)3 + 23 + (3 × 100 × 2) (100 + 2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3

Solution:

998 = 1000 – 2

Using formula, (x – y)3 = x3 – y3 – 3xy(x – y)

(998)3 = (1000 – 2)3

= (1000)3 – 23 – (3 × 1000 × 2) (1000 – 2)

= 1000000000 – 8 – 6000(1000 –  2)

= 1000000000 – 8 – 6000000 + 12000

= 994011992

### Question 8. Factorise each of the following:

(i) 8a3 + b3 + 12a2b + 6ab2

Solution:

8a3 + b3 +12a2b + 6ab2 can also be written as (2a)3 + b3 + 3(2a)2b + 3(2a)(b)2

8a3 + b3 + 12a2b + 6ab2 = (2a)3 + b3 + 3(2a)2b + 3(2a)(b)2

Formula used, (x + y)3 = x3 + y3 + 3xy(x + y)

= (2a + b)3

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

(ii) 8a3 – b3 – 12a2b + 6ab2

Solution:

8a3 – b3 − 12a2b + 6ab2 can also be written as (2a)3– b3 – 3(2a)2b + 3(2a)(b)2

8a3 – b3 − 12a2b + 6ab2 = (2a)3 – b3 – 3(2a)2b + 3(2a)(b)2

formula used, (x – y)3 = x3 – y3 – 3xy(x – y)

= (2a – b)3

= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a3 – 135a + 225a2

Solution:

27 – 125a3 – 135a +225a2 can be also written as 33 – (5a)3 – 3(3)2(5a) + 3(3)(5a)2

27 – 125a3 – 135a + 225a2 = 33 – (5a)3 – 3(3)2(5a) + 3(3)(5a)2

Formula used, (x – y)3 = x3 – y3 – 3xy(x – y)

= (3 – 5a)3

= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a3 – 27b3 – 144a2b + 108ab2

Solution:

64a3 – 27b3 – 144a2b + 108ab2 can also be written as (4a)3 – (3b)3 – 3(4a)2(3b) + 3(4a)(3b)2

64a3 – 27b3 – 144a2b + 108ab2 = (4a)3 – (3b)3 – 3(4a)2(3b) + 3(4a)(3b)2

Formula used, (x – y)3 = x3 – y3 – 3xy(x – y)

= (4a – 3b)3

= (4a – 3b) (4a – 3b) (4a – 3b)

(v) 7p3 − p2 p

Solution:

27p3 – − ( ) p2 + ( )p can also be written as (3p)3 – – 3(3p)2( ) + 3(3p)( )2

27p3 – ( ) − ( ) p2 + ( )p = (3p)3 – ( )3 – 3(3p)2( ) + 3(3p)( )2

Formula used, (x – y)3 = x3 – y3 – 3xy(x – y)

= (3p – )3

= (3p – ) (3p – ) (3p – )

### Chapter 2 Polynomials – Exercise 2.5 | Set 2

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