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Class 9 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.4

Last Updated : 03 Apr, 2024
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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[Tex]\frac{3}{2}     [/Tex]) (y2 – [Tex]\frac{3}{2}     [/Tex])

Solution:

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

[So, a = y2 and b = [Tex]\frac{3}{2}     [/Tex]]

(y 2[Tex]\frac{3}{2}     [/Tex]) (y2 – [Tex]\frac{3}{2}     [/Tex]) = (y2)2 – ([Tex]\frac{3}{2}     [/Tex])^2

= y 4 – [Tex]\frac{9}{4}[/Tex]

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 – [Tex]\frac{y^2}{100}[/Tex]

Solution:

x2 – [Tex]\frac{y^2}{100}     [/Tex] = x2 – [Tex](\frac{y}{10})^2[/Tex]

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

Then, 

a = x

b = [Tex]\frac{y}{10}[/Tex]

= (x – [Tex]\frac{y}{10}     [/Tex]) (x + [Tex]\frac{y}{10}     [/Tex])

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) ([Tex]\frac{1}{4}     [/Tex]a – [Tex]\frac{1}{2}     [/Tex]b + 1)2

Solution:

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

Then, 

x = [Tex]\frac{1}{4}     [/Tex]a

y = [Tex]\frac{-1}{2}     [/Tex]b

z = 1

([Tex]\frac{1}{4}     [/Tex]a – ([Tex]\frac{-1}{2}     [/Tex])b + 1)2 = [[Tex]\frac{1}{4}     [/Tex]a]2 + [[Tex]\frac{-1}{2}     [/Tex]b]2 + 12 + [2 x [Tex]\frac{1}{4}     [/Tex]}a x [Tex]\frac{-1}{2}     [/Tex]b] + [2 x[Tex]\frac{-1}{2}     [/Tex]b x 1] + [2 x 1 x  [Tex]\frac{1}{4}     [/Tex]a]

[Tex]\frac{1}{16}     [/Tex]a2[Tex]\frac{1}{4}     [/Tex]b2 + 1 – [Tex]\frac{1}{4}     [/Tex]ab – b + [Tex]\frac{1}{2}     [/Tex]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) ([Tex]\frac{3}{2}     [/Tex]x + 1)3

Solution:

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

([Tex]\frac{3}{2}     [/Tex]x+ 1)3 = ([Tex]\frac{3}{2}     [/Tex]x)3 + 13 + (3 × [Tex]\frac{3}{2}     [/Tex]x × 1) ([Tex]\frac{3}{2}     [/Tex]x + 1)

[Tex]\frac{27}{8}     [/Tex]x3 + 1 + [Tex]\frac{27}{4}     [/Tex]x2[Tex]\frac{9}{2}     [/Tex]x

[Tex]\frac{27}{8}x^3  + \frac{27}{4}x^2 + \frac{9}{2}x +1 [/Tex]

(iv) (x − [Tex]\frac{2}{3}     [/Tex]y)3

Solution:

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

(x − [Tex]\frac{2}{3}     [/Tex]y)3 = x3 − [[Tex]\frac{2}{3}     [/Tex]y]3 – 3(x) [Tex]\frac{2}{3}     [/Tex]y[x − [Tex]\frac{2}{3}     [/Tex]y]

= x3[Tex]\frac{8}{27}     [/Tex]y3 – 2x2y + [Tex]\frac{4}{3}     [/Tex]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[Tex]\frac{1}{216}     [/Tex] − [Tex]\frac{9}{2}     [/Tex] p2[Tex]\frac{1}{4}     [/Tex]p

Solution:

27p3 – [Tex]\frac{1}{216}     [/Tex] − ([Tex]\frac{9}{2}     [/Tex]) p2 + ([Tex]\frac{1}{4}[/Tex])p can also be written as (3p)3 – [Tex](\frac{1}{6})^3[/Tex] – 3(3p)2([Tex]\frac{1}{6}     [/Tex]) + 3(3p)([Tex]\frac{1}{6}     [/Tex])2

27p3 – ([Tex]\frac{1}{216}[/Tex]) − ([Tex]\frac{9}{2}[/Tex]) p2 + ([Tex]\frac{1}{4}[/Tex])p = (3p)3 – ([Tex]\frac{1}{6}[/Tex])3 – 3(3p)2([Tex]\frac{1}{6}[/Tex]) + 3(3p)([Tex]\frac{1}{6}[/Tex])2

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

= (3p – [Tex]\frac{1}{6}[/Tex])3

= (3p – [Tex]\frac{1}{6}[/Tex]) (3p – [Tex]\frac{1}{6}[/Tex]) (3p – [Tex]\frac{1}{6}[/Tex])

Question 9. Verify:

(i) x3 + y3 = (x + y) (x2 – xy + y2)

Solution:

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

x3 + y3 = (x + y)3 – 3xy(x + y)

x3 + y3 = (x + y) [(x + y)2 – 3xy]

x3 + y3 = (x + y) [(x2 + y2 + 2xy) – 3xy]

Therefore, x3 + y3 = (x + y) (x2 + y2 – xy)

(ii) x3 – y3 = (x – y) (x2 + xy + y2)  

Solution:

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

x3 − y3 = (x – y)3 + 3xy(x – y)

x3− y3 = (x – y) [(x – y)2 + 3xy]

 x3 − y3 = (x – y) [(x2 + y2 – 2xy) + 3xy]

Therefore, x3 + y3 = (x – y) (x2 + y2 + xy)

Question 10. Factorize each of the following:

(i) 27y3 + 125z3

Solution:

27y3 + 125z3 can also be written as (3y)3 + (5z)3

27y3 + 125z3 = (3y)3 + (5z)3

Formula x3 + y3 = (x + y) (x2 – xy + y2)

27y3 + 125z3 = (3y)3 + (5z)3

= (3y + 5z) [(3y)2 – (3y)(5z) + (5z)2]

= (3y + 5z) (9y2 – 15yz + 25z2)

(ii) 64m3 – 343n3

Solution:

64m3 – 343n3 can also be written as (4m)3 – (7n)3

64m3 – 343n3 = (4m)3 – (7n)3

Formula x3 – y3 = (x – y) (x2 + xy + y2)

64m3 – 343n3 = (4m)3 – (7n)3

= (4m – 7n) [(4m)2 + (4m)(7n) + (7n)2]

Question 11. Factorise: 27x3 + y3 + z3 – 9xyz  

Solution:

27x3 + y3 + z3 – 9xyz can also be written as (3x)3 + y3 + z3 – 3(3x)(y)(z)

27x3 + y3 + z3 – 9xyz  = (3x)3 + y3 + z3 – 3(3x)(y)(z)

Formula, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy  – yz – zx)

27x3 + y3 + z3 – 9xyz  = (3x)3 + y3 + z3 – 3(3x)(y)(z)

= (3x + y + z) [(3x)2 + y2 + z2 – 3xy – yz – 3xz]

= (3x + y + z) (9x2 + y2 + z2 – 3xy – yz – 3xz)

Question 12. Verify that: x3 + y3 + z3 – 3xyz  = (1/2) (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]

Solution:

Formula, x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – xz)

Multiplying by 2 and dividing by 2

= (1/2) (x + y + z) [2(x2 + y2 + z2 – xy – yz – xz)]

= (1/2) (x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2xz)

= (1/2) (x + y + z) [(x2 + y2 − 2xy) + (y2 + z2 – 2yz) + (x2 + z2 – 2xz)]

= (1/2) (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]

Therefore, x3 + y3 + z3 – 3xyz  = (1/2) (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]

Question 13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Solution:

Formula, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – xz)

Given, (x + y + z) = 0,

Then, x3 + y3 + z3 – 3xyz = (0) (x2 + y2 + z2 – xy – yz – xz)

x3 + y3 + z3 – 3xyz = 0

Therefore, 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

Solution:

Let,

x = −12

y = 7

z = 5

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.

and we have −12 + 7 + 5 = 0

Therefore, (−12)3 + (7)3 + (5)3 = 3xyz

= 3 × -12 × 7 × 5

= -1260

(ii) (28)3 + (−15)3 + (−13)3

Solution:

Let, 

x = 28

y = −15

z = −13

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.

and we have, x + y + z = 28 – 15 – 13 = 0

Therefore, (28)3 + (−15)3 + (−13)3 = 3xyz

= 3 (28) (−15) (−13)

= 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

Solution:

Using splitting the middle term method,

25a2 – 35a + 12

25a2 – 35a + 12 = 25a2 – 15a − 20a + 12

= 5a(5a – 3) – 4(5a – 3)

= (5a – 4) (5a – 3)

Possible expression for length & breadth is  = (5a – 4)  & (5a  – 3)

(ii) Area : 35y2 + 13y – 12

Solution:

Using the splitting the middle term method,

35y2 + 13y – 12 = 35y2 – 15y + 28y – 12

= 5y(7y – 3) + 4(7y – 3)

= (5y + 4) (7y – 3)

Possible expression for length  & breadth is = (5y + 4) & (7y – 3)

Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?  

(i) Volume : 3x2 – 12x

Solution:

3x2 – 12x can also be written as 3x(x – 4) 

= (3) (x) (x – 4) 

Possible expression for length, breadth & height = 3, x & (x – 4)

(ii) Volume: 12ky2 + 8ky – 20k

Solution:

12ky2 + 8ky – 20k can also be written as 4k (3y2 + 2y – 5)

12ky2 + 8ky– 20k = 4k(3y2 + 2y – 5)

Using splitting the middle term method.

= 4k (3y2 + 5y – 3y – 5)

= 4k [y(3y + 5) – 1(3y + 5)]

= 4k (3y + 5) (y – 1)

Possible expression for length, breadth & height= 4k, (3y + 5) & (y – 1)



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