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RD Sharma Class 9 Solutions Chapter 5 Exercise 5.1
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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
= x2 – 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 )( 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 a 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)
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