NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5.

• Polynomials Class 9 Ex 2.1
• Polynomials Class 9 Ex 2.2
• Polynomials Class 9 Ex 2.3
• Polynomials Class 9 Ex 2.4
• Polynomials Class 9 Ex 2.5

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 2
Chapter Name	Polynomials
Exercise	Ex 2.5
Number of Questions Solved	16
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 1.
Use suitable identities to find the following products:
(i) (x+4)(x+10)
(ii) (x+8)(x-10)
(iii) (3x+4)(3x+2x)
(iv) ()( )
(v) (3-2x)(3+2x)
Solution:

Question 2.
Evaluate the following products without multiplying directly:
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
Solution:
(i) 103 x 107 = (100 + 3) (100 + 7)
= (100)² + (3 + 7) (100) + 3 x 7
= 100 x 100 + (10)(100) + 21
= 10000 +1000 + 21 = 11021

(ii) 95 x 96 = (100 – 5) (100 – 4)
= (100)² + (-5 – 4)(100) + (-5)(-4) – 100 x 100 + (-9)(100) + 20
= 10000 – 900 + 20 = 9120

(iii) 104 x 96 = (100 + 4)(100 – 4)
= (100)² – (4)²
= 10000 -16
= 9984

Question 3.
Factorise the following using appropriate identities :
(i) 9x² + 6xy + y²
(ii) 4y² – 4y +1
(iii) x²–
Solution:

Question 4.
Expand each of the following, using suitable identities :
(i) (x + 2y + 4z)²
(ii) (2x -y +z)²
(iii) (-2x + 3y + 2z)²
(iv) (3a – 7b -c)²
(v) (-2x + 5y – 3z)²
(vi) ²
Solution:


Question 5.
Factorise:
(i) 4x² + 9y² + I62² + 12xy – 24yz -16xz
(ii) 2x²+ y² + 82² – 2xy + 4yz – 8×2
Solution:
(i) 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

(ii) 2x² + y² + 8z² – 2xy + 4yz-8xz
= (-ax)² + (y)² + (2z)² + (2 – x) (y) + 2(y) (2z) + 2(2z)(-x)
= (-x + y + 2z)²

Question 6.
Write the following cubes in expanded form :
(i) (2x+1)³
(ii) (2a-3b)³
(iii) ³
(iv) ³
Solution:


Question 7.
Evaluate the following using suitable identities :
(i) (99)³
(ii) (102)³
(iii) (998)³
Solution:

Question 8.
Factorise each of the following:
(i) 8a³ + b³ + 12a²6 + 6ab²
(ii) 8a³ – b³ -12a²6 + 6a6²
(iii) 27 – 125a³ – 135 a + 225 a² 64a³
(iv)
Solution:


Question 9.
Verify:
(1) x³ + y³ = Or + y)(x²-xy + y²)
(ii) x³-y³ = (x-y)(x² + xy + y²)
Solution:
(i) We know that,
(x + y)³ = x³ + y³ + 3xy(x + y)
=> x³ + y³ = (x + y)³ -3xy(x + y)
= (x + y)[(x + y)² -3xy]
= (x + y) [x² + y² + 2xy – 3x] = (x + y)[x² + y² – xy]
= RHS  Hence proved.

(ii) We know that, (x – y)³ = x³ – y³ -3xy(x – y)
=>x³ – y³ = (x – y)³ + 3xy(x – y)
= (x-y)[(x – y)² +3xy]
= (x -y)[x² + y² -2xy + 3xy]
= (x — y)[x² + y² + xy]
= RHS  Hence proved.

Question 10.
Factorise each of the following:
(i) 27y³ + 125z³
(ii) 64m³ -343n³ [Hint: See question 9]
Solution:
(i) 27 y³ +125z³ = (3y)³ + (5z)³
= (3y + 5z)[(3y)² – (3y)(5z) + (5z)²]
= (3y + 5z) (9 y² – 15yz + 25z²)

(ii) 64m³ -343n³ = (4m)³ -(7n)³
= (4m-7n)[(4m)² + (4m)(7n) + (7n)²]
= (4m – 7n)[16m² + 28mn + 49n²]

Question 11.
Factorise :
27x³ + y³ + z³ – 9xyz
Solution:
27x³ +y³ +z³ -9xyz = (3x)³ + y³ +z³ -3(3x)(y)(z)
= (3x + y + z)[(3x)² + y² + z² – (3x)y – yz -z(3x)]
= (3x + y + z)(9x² + y² + z² -3xy – yz -3zx)

Question 12.
Verify that
x³ + y³ +z³ -3xyz = (x + y + z)[(x -y)² +(y-z)² +(z-x)²]
Solution:
We have, x³ + y³ + z³ – 3xyz
= (x + y + z) [x² + y² + z² – xy – yz – zx]
= (x + y+ z)[2x² +2y² +2z² -2xy-2yz -2zx]
= (x + y + z)[x² + x² + y² + y² + z² + z² -2xy-2yz-2zx]
= (x + y + z)[x² + y² – 2xy + y² + z² -2yz + z² + x² – 2zx]
= (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:
We know that,
x³ +y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz-zx)
= 0(x² + y² + z² – xy- yz-zx) (∵ x + y + z = 0 given)
= 0
=> x³ + y³ + z³ = 3xyz        Hence proved.

Question 14.
Without actually calculating the cubes, find the value of each of the following: 
(i) (-12)³ + (7)³ + (5)³
(ii) (28)³ + (-15)³ + (-13)³.
Solution:
(i) We know that, x³ + y³ + z³ – 3xyz
= (x + y + z)(x² + y² + z² – xy – yz – zx)
Also, we know that, if
x + y + z = 0
Then, x³ + y³ +z³ = 3 xyz
Given expression is (-12)³ + (7)³ + (5)³.
Here,            -12 + 7 + 5=0
∴ (-12)³ + (7)³ + (5)³ = 3 x (-12) x 7 x 5 = -1260

(ii) Given expression is (28)³ + (-15)³ + (-13)³
Here, 28 + (-15) + (-13) = 28 -15-13 = 0
∴  (28)³ + (-15)³ + (-13)³ = 3 x (28) x (-15) x (-13) = 16380

Question 15.
Give possible expressions for the length and the breadth of each of the following rectangles, in which their areas are given :
(i) Area = 25a² – 35a + 12
(ii) Area = 35y² + 13y – 12.
Solution:
(i) We have,
Area of rectangle = 25a² – 35a +12 [by splitting the middle term]
= 25a² -20a-15a+12
= 5a(5a – 4) – 3(5a – 4)
= (5a-4)(5a-3)
Hence, possible expression for length = (5a – 3) and possible expression for breadth = (5a – 4)

(ii) We have,
Area of rectangle = 35y+ 13y -12
= 35y² + (28 – 15)y -12
= 35y² + 28y-15y-12
= (35y² +28y)-(15y + 12)
= 7y(5y + 4) – 3(5y + 4)
= (7y – 3) (5y + 4)

Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume = 3x² – 12x
(ii) Volume = 12ky² + 8ky – 20k.
Solution:
(i) We have,
Volume of cuboid = 3x² -12x = 3x(x – 4)
So, the possible expressions for the dimensions of the cuboids are 3, x and x – 4.
[∵ volume of cuboid = length x breadth x height]

(ii) We have,
Volume of cuboid = 12ky² + 8ky – 20k = 12ky² + (20 -12)ky – 20k [by splitting the middle term]
= 12ky² + 20 ky – 12ky – 20 k = 4ky(3y + 5) – 4k(3y + 5) = (3y + 5)(4ky – 4k)
= (3y + 5)4k(y -1) = 4k(3y + 5)(y -1)
So, the possible expressions for the dimensions of the cuboid are 4 k, 3y + 5 and y -1.

We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5, drop a comment below and we will get back to you at the earliest.