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

• Last Updated : 26 Jul, 2022

### 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 Ã— 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 = âˆ’10]

(x + 8) (x – 10) = x2 + (8 + (-10) )x + (8 Ã— (-10))

= x2 + (8 – 10) x – 80

= x2 âˆ’ 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 = âˆ’5]

(3x + 4) (3x âˆ’ 5) = (3x)2 + [4 + (-5)]3x + 4 Ã— (-5)

= 9x2 + 3x (4 – 5) – 20

= 9x2 â€“ 3x â€“ 20

(iv) (y2) (y2 – )

Solution:

Using formula, (a + b) (a â€“ b) = a2 â€“ b2

[So, a = y2 and b = ]

(y 2) (y2 â€“ ) = (y2)2 â€“ ()^2

= y 4 â€“

### Question 2. Evaluate the following products without multiplying directly:

(i) 103 Ã— 107

Solution:

103 Ã— 107 = (100 + 3) Ã— (100 + 7)

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

Then,

x = 100

a = 3

b = 7

So, 103 Ã— 107 = (100 + 3) Ã— (100 + 7)

= (100)2 + (3 + 7)100 + (3 Ã— 7)

= 10000 + 1000 + 21

= 11021

(ii) 95 Ã— 96

Solution:

95 Ã— 96 = (100 – 5) Ã— (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 Ã— 96 = (100 – 5) Ã— (100 – 4)

= (100)2 – 100 (5+4) + (5 Ã— 4)

= 10000 – 900 + 20

= 9120

(iii) 104 Ã— 96

Solution:

104 Ã— 96 = (100 + 4) Ã— (100 â€“ 4)

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

Then,

a = 100

b = 4

So, 104 Ã— 96 = (100 + 4) Ã— (100 â€“ 4)

= (100)2 â€“ (4)2

= 10000 â€“ 16

= 9984

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

(i) 9x2 + 6xy + y2

Solution:

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

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

Then,

a = 3x

b = y

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

= (3x + y)2

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

(ii) 4y2 âˆ’ 4y + 1

Solution:

4y2 âˆ’ 4y + 1 = (2y)2 â€“ (2 Ã— 2y Ã— 1) + 1

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

Then,

a = 2y

b = 1

= (2y â€“ 1)2

= (2y â€“ 1) (2y â€“ 1)

(iii)  x2 –

Solution:

x2 â€“  = x2 â€“

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

Then,

a = x

b =

= (x â€“ ) (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 Ã— x Ã— 2y) + (2 Ã— 2y Ã— 4z) + (2 Ã— 4z Ã— x)

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

(ii) (2x âˆ’ y + z)2

Solution:

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

Then,

x = 2x

y = âˆ’y

z = z

(2x âˆ’ y + z)2 = (2x)2 + (âˆ’y)2 + z2 + (2 Ã— 2x Ã— âˆ’y) + (2 Ã— âˆ’y Ã— z) + (2 Ã— z Ã— 2x)

= 4x2 + y2 + z2 â€“ 4xy â€“ 2yz + 4xz

(iii) (âˆ’2x + 3y + 2z)2

Solution:

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

Then,

x = âˆ’2x

y = 3y

z = 2z

(âˆ’2x + 3y + 2z)2 = (âˆ’2x)2 + (3y)2 + (2z)2 + (2 Ã—âˆ’2x Ã— 3y) + (2 Ã—3y Ã— 2z) + (2 Ã—2z Ã— âˆ’2x)

= 4x2 + 9y2 + 4z2 â€“ 12xy + 12yzâ€“ 8xz

(iv) (3a â€“ 7b â€“ c)2

Solution:

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

Then,

x = 3a

y = â€“ 7b

z = â€“ c

(3a â€“ 7b â€“  c)2 = (3a)2 + (â€“ 7b)2 + (â€“ c)2 + (2 Ã— 3a Ã— â€“ 7b) + (2 Ã— â€“7b Ã— â€“c) + (2 Ã— â€“c Ã— 3a)

= 9a2 + 49b2 + c2 â€“ 42ab + 14bc â€“ 6ca

(v) (â€“2x + 5y â€“ 3z)2

Solution:

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

Then,

x = â€“2x

y = 5y

z = â€“ 3z

(â€“2x + 5y â€“ 3z)2 = (â€“2x)2 + (5y)2 + (â€“3z)2 + (2 Ã— â€“2x Ã— 5y) + (2 Ã—  5y Ã— â€“ 3z) + (2 Ã— â€“3z Ã— â€“2x)

= 4x2 + 25y2 + 9z2 â€“ 20xy â€“ 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 xb x 1] + [2 x 1 x  a]

a2b2 + 1 – ab – b + a

### Question 5. Factorize:

(i) 4x2 + 9y2 + 16z2 + 12xy â€“ 24yz â€“ 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 â€“ 24yz â€“ 16xz = (2x)2 + (3y)2 + (âˆ’4z)2 + (2 Ã— 2x Ã— 3y) + (2 Ã— 3y Ã— âˆ’4z) + (2 Ã— âˆ’4z Ã— 2x)

= (2x + 3y â€“ 4z)2

= (2x + 3y â€“ 4z) (2x + 3y â€“ 4z)

(ii) 2x2 + y2 + 8z2 â€“ 2âˆš2xy + 4âˆš2yz â€“ 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 â€“ 2âˆš2xy + 4âˆš2yz â€“ 8xz

= (-âˆš2x)2 + (y)2 + (2âˆš2z)2 + (2 Ã— -âˆš2x Ã— y) + (2 Ã— y Ã— 2âˆš2z) + (2 Ã— 2âˆš2 Ã— âˆ’âˆš2x)

= (âˆ’âˆš2x + y + 2âˆš2z)2

= (âˆ’âˆš2x + y + 2âˆš2z) (âˆ’âˆš2x + y + 2âˆš2z)

### 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 Ã— 2x Ã—1) (2x + 1)

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

= 8x3 + 12x2 + 6x + 1

(ii) (2a âˆ’ 3b)3

Solution:

Using formula, (x â€“ y)3 = x3 â€“ y3 â€“ 3xy(x â€“ y)

(2a âˆ’ 3b)3 = (2a)3 âˆ’ (3b)3 â€“ (3 Ã— 2a Ã— 3b) (2a â€“ 3b)

= 8a3 â€“ 27b3 â€“ 18ab(2a â€“ 3b)

= 8a3 â€“ 27b3 â€“ 36a2b + 54ab2

(iii) (x + 1)3

Solution:

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

(x+ 1)3 = (x)3 + 13 + (3 Ã— x Ã— 1) (x + 1)

x3 + 1 + x2x

(iv) (x âˆ’ y)3

Solution:

Using formula, (x â€“ y)3 = x3 â€“ y3 â€“ 3xy(x â€“ y)

(x âˆ’ y)3 = x3 âˆ’ [y]3 – 3(x) y[x âˆ’ y]

= x3y3 – 2x2y + xy2

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

(i) (99)3

Solution:

99 = 100 â€“ 1

Using formula, (x â€“ y)3 = x3 â€“ y3 â€“ 3xy(x â€“ y)

(99)3 = (100 â€“ 1)3

= (100)3 â€“ 13 â€“ (3 Ã— 100 Ã— 1) (100 â€“ 1)

= 1000000 â€“ 1 â€“ 300(100 â€“ 1)

= 1000000 â€“ 1 â€“ 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 Ã— 100 Ã— 2) (100 + 2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3

Solution:

998 = 1000 â€“ 2

Using formula, (x â€“ y)3 = x3 â€“ y3 â€“ 3xy(x â€“ y)

(998)3 = (1000 â€“ 2)3

= (1000)3 â€“ 23 â€“ (3 Ã— 1000 Ã— 2) (1000 â€“ 2)

= 1000000000 â€“ 8 â€“ 6000(1000 â€“  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 â€“ b3 â€“ 12a2b + 6ab2

Solution:

8a3 â€“ b3 âˆ’ 12a2b + 6ab2 can also be written as (2a)3â€“ b3 â€“ 3(2a)2b + 3(2a)(b)2

8a3 â€“ b3 âˆ’ 12a2b + 6ab2 = (2a)3 â€“ b3 â€“ 3(2a)2b + 3(2a)(b)2

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

= (2a â€“ b)3

= (2a â€“ b) (2a â€“ b) (2a â€“ b)

(iii) 27 â€“ 125a3 â€“ 135a + 225a2

Solution:

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

27 â€“ 125a3 â€“ 135a + 225a2 = 33 â€“ (5a)3 â€“ 3(3)2(5a) + 3(3)(5a)2

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

= (3 â€“ 5a)3

= (3 â€“ 5a) (3 â€“ 5a) (3 â€“ 5a)

(iv) 64a3 â€“ 27b3 â€“ 144a2b + 108ab2

Solution:

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

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

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

= (4a â€“ 3b)3

= (4a â€“ 3b) (4a â€“ 3b) (4a â€“ 3b)

(v) 7p3 â€“ âˆ’  p2p

Solution:

27p3 â€“  âˆ’ () p2 + ()p can also be written as (3p)3 â€“  â€“ 3(3p)2() + 3(3p)()2

27p3 â€“ () âˆ’ () p2 + ()p = (3p)3 â€“ ()3 â€“ 3(3p)2() + 3(3p)()2

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

= (3p â€“ )3

= (3p â€“ ) (3p â€“ ) (3p â€“ )

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

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