Class 9 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.4

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 xb 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 – )

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)