NCERT Solutions for Class 9 Maths Polynomials: Exercise 5

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)

NCERT Solutions Class 9 Maths Chapter 2 Polynomials is curated by a team of professionals at GFG to help students in their academic journey. In this article, we have provided the answers to all the questions in the NCERT textbook. All of the problems in this chapter’s exercise from the NCERT textbook for Class 9 are covered in the NCERT Solutions for Class 9 Maths.

The topics covered in the Polynomials chapter in Class 9 are Polynomials, Types of polynomials, Operations on polynomials, Degree of a polynomial, Factorization of polynomials, Factor Theorem, Remainder theorem, Algebraic identities and Application problems are the important Chapters in Class 9 Maths.