RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1 (Updated for 2024)

RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1

RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1: You can download the Free PDF of RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1 using the link mentioned in our blog. You can easily learn the concepts of Class 9 Maths with this amazing guide. All your Maths assignments and tests will be covered in this. To know more about the RD Sharma Solutions Class 9 Maths, read the whole blog.

Download RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1 PDF

RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1

 


Access answers of RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1

Exercise 5.1 Page No: 5.9

Question 1: Factorize x3 + x – 3x2 – 3

Solution:

x3 + x – 3x2 – 3

Here x is common factor in x3 + x and – 3 is common factor in – 3x2 – 3

x3 – 3x2 + x – 3

x2 (x – 3) + 1(x – 3)

Taking ( x – 3) common

(x – 3) (x2 + 1)

Therefore x3 + x – 3x2 – 3 = (x – 3) (x2 + 1)

Question 2: Factorize a(a + b)3 – 3a2b(a + b)

Solution:

a(a + b)3 – 3a2b(a + b)

Taking a(a + b) as common factor

= a(a + b) {(a + b)2 – 3ab}

= a(a + b) {a2 + b2 + 2ab – 3ab}

= a(a + b) (a2 + b2 – ab)

Question 3: Factorize x(x3 – y3) + 3xy(x – y)

Solution:

x(x3 – y3) + 3xy(x – y)

= x(x – y) (x2 + xy + y2) + 3xy(x – y)

Taking x(x – y) as a common factor

= x(x – y) (x2 + xy + y2 + 3y)

= x(x – y) (x2 + xy + y2 + 3y)

Question 4: Factorize a2x2 + (ax2 + 1)x + a

Solution:

a2x2 + (ax2 + 1)x + a

= a2x2 + a + (ax2 + 1)x

= a(ax2 + 1) + x(ax2 + 1)

= (ax2 + 1) (a + x)

Question 5: Factorize x2 + y – xy – x

Solution:

x2 + y – xy – x

= x– x – xy + y

= x(x- 1) – y(x – 1)

= (x – 1) (x – y)

Question 6: Factorize x3 – 2x2y + 3xy2 – 6y3

Solution:

x3 – 2x2y + 3xy2 – 6y3

= x2(x – 2y) + 3y2(x – 2y)

= (x – 2y) (x2 + 3y2)

Question 7: Factorize 6ab – b2 + 12ac – 2bc

Solution:

6ab – b2 + 12ac – 2bc

= 6ab + 12ac – b2 – 2bc

Taking 6a common from the first two terms and –b from the last two terms

= 6a(b + 2c) – b(b + 2c)

Taking (b + 2c) common factor

= (b + 2c) (6a – b)

Question 8: Factorize (x2 + 1/x2) – 4(x + 1/x) + 6

Solution:

(x2 + 1/x2) – 4(x + 1/x) + 6

= x2 + 1/x2 – 4x – 4/x + 4 + 2

= x2 + 1/x2 + 4 + 2 – 4/x – 4x

= (x2) + (1/x) 2 + (-2)2 + 2x(1/x) + 2(1/x)(-2) + 2(-2)x

As we know, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x+y+z) 2

So, we can write;

= (x + 1/x + (-2 )) 2

or (x + 1/x – 2) 2

Therefore, x2 + 1/x2) – 4(x + 1/x) + 6 = (x + 1/x – 2) 2

Question 9: Factorize x(x – 2) (x – 4) + 4x – 8

Solution:

x(x – 2) (x – 4) + 4x – 8

= x(x – 2) (x – 4) + 4(x – 2)

= (x – 2) [x(x – 4) + 4]

= (x – 2) (x2 – 4x + 4)

= (x – 2) [x2 – 2 (x)(2) + (2) 2]

= (x – 2) (x – 2) 2

= (x – 2)3

Question 10: Factorize ( x + 2 ) ( x2 + 25 ) – 10x2 – 20x

Solution :

( x + 2) ( x2 + 25) – 10x ( x + 2 )

Take ( x + 2 ) as common factor;

= ( x + 2 )( x+ 25 – 10x)

=( x + 2 ) ( x– 10x + 25)

Expanding the middle term of ( x2 – 10x + 25 )

=( x + 2 ) ( x– 5x – 5x + 25 )

=( x + 2 ){ x (x – 5 ) – 5 ( x – 5 )}

=( x + 2 )( x – 5 )( x – 5 )

=( x + 2 )( x – 5 )2

Therefore, ( x + 2) ( x2 + 25) – 10x ( x + 2 ) = ( x + 2 )( x – 5 )2

Question 11: Factorize 2a2 + 2√6 ab + 3b2

Solution:

2a2 + 2√6 ab + 3b2

Above expression can be written as ( √2a )2 + 2 × √2a × √3b + ( √3b)2

As we know, ( p + q ) 2 = p+ q2 + 2pq

Here p = √2a and q = √3b

= (√2a + √3b )2

Therefore, 2a2 + 2√6 ab + 3b2 = (√2a + √3b )2

Question 12: Factorize (a – b + c)2 + (b – c + a) 2 + 2(a – b + c) (b – c + a)

Solution:

(a – b + c)2 + ( b – c + a) 2 + 2(a – b + c) (b – c + a)

{Because p+ q2 + 2pq = (p + q) 2}

Here p = a – b + c and q = b – c + a

= [a – b + c + b- c + a]2

= (2a)2

= 4a2

Question 13: Factorize a2 + b+ 2( ab+bc+ca )

Solution:

a2 + b+ 2ab + 2bc + 2ca

As we know, p+ q2 + 2pq = (p + q) 2

We get,

= ( a+b)2 + 2bc + 2ca

= ( a+b)2 + 2c( b + a )

Or ( a+b)2 + 2c( a + b )

Take ( a + b ) as a common factor;

= ( a + b )( a + b + 2c )

Therefore, a2 + b+ 2ab + 2bc + 2ca = ( a + b )( a + b + 2c )

Question 14: Factorize 4(x-y) 2 – 12(x – y)(x + y) + 9(x + y)2

Solution :

Consider ( x – y ) = p, ( x + y ) = q

= 4p2 – 12pq + 9q2

Expanding the middle term, -12 = -6 -6 also 4× 9=-6 × -6

= 4p2 – 6pq – 6pq + 9q2

=2p( 2p – 3q ) -3q( 2p – 3q )

= ( 2p – 3q ) ( 2p – 3q )

= ( 2p – 3q )2

Substituting back p = x – y and q = x + y;

= [2( x-y ) – 3( x+y)]= [ 2x – 2y – 3x – 3y ] 2

= (2x-3x-2y-3y ) 2

=[ -x – 5y] 2

=[( -1 )( x+5y )] 2

=( x+5y ) 2

Therefore, 4(x-y) 2 – 12(x – y)(x + y) + 9(x + y)2 = ( x+5y )2

Question 15: Factorize a2 – b+ 2bc – c2

Solution :

a2 – b+ 2bc – c2

As we know, ( a-b)2 = a+ b– 2ab

= a– ( b – c) 2

Also we know, a– b= ( a+b)( a-b)

= ( a + b – c )( a – ( b – c ))

= ( a + b – c )( a – b + c )

Therefore, a2 – b+ 2bc – c=( a + b – c )( a – b + c )

Question 16: Factorize a2 + 2ab + b2 – c2

Solution:

a2 + 2ab + b2 – c2

= (a2 + 2ab + b2) – c2

= (a + b)2 – (c) 2

We know, a2 – b2 = (a + b) (a – b)

= (a + b + c) (a + b – c)

Therefore a2 + 2ab + b2 – c= (a + b + c) (a + b – c)


This is the complete blog on RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1. To know more about the CBSE Class 9 Maths exam, ask in the comments. 

FAQs on RD Sharma Class 9 Solutions for Chapter 5 Exercise 5.1

How many questions are there in RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1?

There are 36 questions in RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1.

From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 5 Exercise 5.1?

You can find the download link from the above blog.

How much does it cost to download the PDF of RD Sharma Solutions Class 9 Maths Chapter 5 Exercise 5.1

You can download it for free.

Leave a Comment

Top 10 Professional Courses With High-Paying Jobs 2024 Top 8 Online MCA Colleges in India 2024 Skills You Will Gain from an Online BBA Programme How to stay motivated during distance learning Things to know before starting with first year of medical school