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

In Class 9, Mathematics students are introduced to the fundamental concept of the polynomials in Chapter 2 of the NCERT textbook. This chapter explores the nature, properties, and applications of polynomials. Exercise 2.5 in Set 1 focuses on solving the problems related to polynomial expressions which helps in understanding their behavior and simplification techniques. Mastering this exercise is crucial as it builds the foundation for the more advanced algebraic concepts and problem-solving skills.

Polynomials

The Polynomials are algebraic expressions consisting of variables and coefficients connected by addition, subtraction, and multiplication operations but not division by a variable. In Class 9, polynomials are studied to understand their degree, roots, and various types. This foundational knowledge is essential as polynomials are widely used in algebra, calculus, and various real-life applications. Exercise 2.5 helps reinforce these concepts through practice problems enhancing students' problem-solving abilities and their grasp of the polynomial operations.

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 – )

Chapter 2 Polynomials - Exercise 2.5 | Set 2

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