RD Sharma class 9 mathematics chapter 5 exercise 5.1 Factorization of Algebraic Expressions is provided here. In this chapter, we will solve different types of algebraic expressions like monomials, binomials, trinomials, and polynomials using identities. Understand these concepts and have in-depth knowledge through some of the solved RD Sharma solutions for 9th standard. BYJU’S experts have prepared RD Sharma Class 9 chapter 5 solutions as per latest CBSE syllabus for class 9. Students can download RD Sharma exercise 5.1 by clicking on the link below.
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Access Answers to Maths RD Sharma Class 9 Chapter 5 Factorization of Algebraic Expressions Exercise 5.1 Page number 5.9
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)
= 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
= x2 – x – xy + y
= x(x- l) – 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 first two terms and –b from 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 )( x2 + 25 – 10x)
=( x + 2 ) ( x2 – 10x + 25)
Expanding the middle term of ( x2 – 10x + 25 )
=( x + 2 ) ( x2 – 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 = p2 + 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 p2 + 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 + b2 + 2( ab+bc+ca )
Solution:
a2 + b2 + 2ab + 2bc + 2ca
As we know, p2 + 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 common factor;
= ( a + b )( a + b + 2c )
Therefore, a2 + b2 + 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)]2 = [ 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 – b2 + 2bc – c2
Solution :
a2 – b2 + 2bc – c2
As we know, ( a-b)2 = a2 + b2 – 2ab
= a2 – ( b – c) 2
Also we know, a2 – b2 = ( a+b)( a-b)
= ( a + b – c )( a – ( b – c ))
= ( a + b – c )( a – b + c )
Therefore, a2 – b2 + 2bc – c2 =( 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 – c2 = (a + b + c) (a + b – c)
Access other exercise solutions of Class 9 Maths Chapter 5 Factorization of Algebraic Expressions
RD Sharma Solutions for Class 9 Maths Chapter 5 Exercise 5.1
Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.1 is based on the following topics:
- Factorisation by taking out the common factors
- Factorisation by grouping the terms
- Factorisation by making a perfect square
- Factorisation by the difference of two squares
- Factorisation of quadratic polynomials by splitting the middle term