RD Sharma Solutions Class 9 Factorization Of Algebraic Expressions Exercise 5.1

RD Sharma Solutions Class 9 Chapter 5 Exercise 5.1

RD Sharma Class 9 Solutions Chapter 5 Ex 5.1 Free Download

RD Sharma Solutions Class 9 Chapter 5 Ex 5.1

Q1 . x+ x – 3x– 3

Solution :

Take out  x as common factor in x3+ x;

x ( x2+1 ) – 3x2-3

Take out  – 3 as common factor in -3x2-3;

x( x2+1)-3( x2+1 )

Now , taking ( x2+1) common;

=( x2+1) (x – 3)

Therefore,  x3+x-3y2-3 = ( x2+1) (x – 3)

 

Q2 . a( a+b)3-3a2b ( a+b )

Solution :

Take out (a + b) as common factor from the two terms;

= (a + b) {a(a + b) ² – 3a ²b}

As we know,   ( a+b)2=a2+b2+2ab

=( a+b ) {a( a2+b2+2ab ) – 3a2b} 

= ( a+b ){ a3+ab2+2a2b-3a2b }

= ( a+b){ a3+ab2-a2b}

=( a+b)p{ a2+b2-ab}

=p( a+b) ( a2+b2-ab)

Therefore, a( a+b)3-3a2b( a+b)=a( a+b) ( a2+b2-ab)

 

Q3 . x( x3-y3)+3xy ( x-y)

Solution:

Expanding x3-y3  using the formula, x3-y3= ( x-y) ( x2+xy+y2)

x( x-y)( x2 +xy+y2)+3xy( x-y)

Take out common factor x( x-y ) from both the terms

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

Therefore,  x ( x3-y3)+3xy( x-y)= x ( x-y) ( x2+xy+y2+3y )

 

Q4 . a2x2+( ax2+1 )x+a

Solution :

Given, a2x2+ ( ax2+1 )x+a

On expanding;

=a2x2+ax3+x+a

Take out the common factor ax2 in ( a2x2+ax3 )and  1in ( x + a )

=ax2( a+x)+1 ( x+a)

=ax2( a+x)+1 ( a+x)

Take out ( a + x ) as common factor from both the terms

= ( a+x)( ax2+1 )

Therefore, a2x2+( ax2+1 )x+a = ( a+x)( ax2+1 )

 

Q5 . x2+y-xy-x

Solution :

On rearranging, we get;

x2-xy-x+y

Take x as common factor from the ( x2-xy)and -1 in(-x+y )

=x( x – y ) – 1 ( x – y )

Take ( x – y ) as common factor;

=( x – y )( x – 1 )

Therefore, x2+y-xy-x = ( x – y )( x – 1 )

 

Q6 . x3-2x2b+3xy2-6y3

Solution :

Take xas common factor from  ( x3-2x2y) and + 3ycommon from  ( 3xy2-6y3)

= x2( x-2y)+3y2( x-2y )

Take ( x – 2y ) as common fcator;

= ( x-2y) ( x2+3y2 )

Therefore, x3-2x2y+3xy2-6y3= ( x-2y)( x2+3y2 )

 

Q7 . 6ab-b2+12ac-2bc

Solution :

Take b as common factor from ( 6ab-b2)and 2c in ( 12ac – 2bc )

=b( 6a – b ) + 2c ( 6a – b )

Take  ( 6a – b ) as common factor;

=( 6a – b )( b + 2c )

Therefore, 6ab-b2+12ac-2bc=( 6a – b )( b + 2c )

 

Q8 . [ x2+1/x2]-4[ x+1/x]+6

Solution :

=x+ 1/x– 4x – 4/x + 4 + 2

=x+ 1/x+ 4 + 2 – 4/x – 4x

=( x2)+ (1/x)2+ ( -2 )2+2 × x × 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

=[ x+1/x -2]2

= [ x+1/x -2 ] [ x+1/x -2]

Therefore, [ x2+1/x2 ]-4 [ x+1/x]+6 = [ x+ 1/x -2 ] [ x+ 1/x -2 ]

 

Q9 . x( x- 2 )( x – 4 ) + 4x – 8

Solution :

=x( x  – 2 )( x – 4 ) + 4( x – 2 )

Take ( x – 2 ) as common factor;

=( x – 2 ){x( x – 4 ) + 4}

=( x – 2 ) { x2-4x+4}

Now expanding the middle term of x2-4x+4

=( x – 2 ){ x2-2x-2x+4}

=( x – 2 ){ x( x – 2 ) -2( x -2 )}

=( x – 2 ){( x – 2 )( x- 2 )}

=( x – 2 )( x – 2 )( x – 2 )

=( x-2)3

Therefore, x( x- 2 )( x – 4 ) + 4x – 8= ( x-2)3

 

Q10 .( 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 )

Therefore, ( x + 2 ) ( x2+25)-10x2– 20x=( x + 2 )( x – 5 )( x – 5 )

 

Q11 . 2a2+2√6ab+3b2

Solution :

= ( √2a )2+2 × √2a × √3b+ ( √3b)2

As we know, ( p+q )2=p2+q2+2pq

= (√2a+√3b )2

=  (√2a+√3b ) (√2a+√3b )

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

 

Q 12 . (a-b+c)2+(b-c+a)2+2(a-b+c) × (b-c+a)

Solution:

Say, ( a – b + c ) = x and ( b – c + a ) = y

⇒ x2+y2+2xy

As we know,   ( a+b)2=a2+b2+2ab

=(x+y)2

Now , substituting  x and y;

=(a–b+c+b-c+a)2

=(2a)2

=4a2

Therefore, (a-b+c)2+(b-c+a)2+2(a-b+c) × (b-c+a) =4a2 

 

Q13 . a2+b2+2( ab+bc+ca )

Solution :

=a2+b2+2ab+2bc+2ca

As we know,  ( p+q)2=p2+q2+2pq

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+2 ( ab+bc+ca) = ( a + b )( a + b + 2c )

 

Q14 . 4(x-y)2-12(x-y)(x+y)+9(x+y)2

Solution :

Say, ( x – y ) = x,( x + y ) = y

= 4x2-12xy+9y2

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

=4x2-6xy-6xy+9y2

=2x( 2x – 3y ) -3y( 2x – 3y )

=( 2x – 3y ) ( 2x – 3y )

=( 2x-3y)2

Substituting  x =  x – y  & y =  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                                               [(-1)2 = 1]

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

 

Q 15 . a2-b2+2bc-c2

Solution :

a2– ( b2-2bc+c2 )

As we know, ( a-b)2=a2+b2-2ab

= a2-( b-c)2

Since,  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 )

 

Q16 . a2+2ab+b2-c2

Solution :

As we know, ( p+q)2=p2+q2+2pq

=( a+b)2-c2

As we know,  p2-q2= ( p+q)( p-q)

=( a + b + c )( a + b – c )

Therefore, a2+2ab+b2-c2 =( a + b + c )( a + b – c )

 

Q 17 . a2+4b2– 4ab-4c2

Solution:

On rearranging;

= a2– 4ab+4b2-4c2

= ( a )2-2 × a × 2b+ ( 2b)2-4c2

As we know,  ( a-b)2=a2+b2-2ab

=( a-2b)2-4c2

= ( a-2b)2-( 2c)2

As we know,   a2-b2=( a+b) ( a-b )

=( a – 2b  – 2c) ( a – 2b  + 2c)

Therefore, a2+4b2– 4ab-4c2 =( a – 2b  – 2c) ( a – 2b  + 2c)

 

Q18 . xy9 -yx9 

Solution:

=xy ( y8-x8 )

=xy ( ( y4)2– ( x4)2 )

As we know, p2-q2= ( p + q )( p – q )

=xy ( y4+x4)( y4-x4)

=xy ( y4+x4) ( ( y2)2– ( x2)2)

As we know,   p2-q2= ( p + q )( p – q )

=xy ( y4+x4)( y2+x2) ( y2-x2)

= xy( y4+x4) ( y2+x2)( y+x) ( y-x )

=xy( x4+y4) ( x2+y2) (x+y) ( -1) ( x-y )

∵  ( y-x)=-1( x-y)

=-xy( x4+y4)( x2+y2)( x+y )( x-y)

Therefore, xy9 -yx9  = -xy( x4+y4)( x2+y2)( x+y )( x-y)

Q 19 . x4+x2y2+y4

Solution:

Adding and subtracting x2y2 to the given equation;

= x4+x2y2+y4+x2y2-x2y2

=x4+2x2y2+y4-x2y2

= ( x2)2+2 × x× y2+( y2)2– ( xy)2

As we know,    ( p+q)2=p2+q2+2pq

= ( x2+y2)2-( xy)2

As we know,   p2-q2= ( p + q )( p – q )

= ( x2+y2+xy)( x2+y2-xy)

Therefore, x4+x2y2+y4= ( x2+y2+xy )( x2+y2-xy)

 

Q20 . x2-y2-4xz+4z2

Solution:

On rearranging the given equation;

= x2-4xz+4z2-y2

= ( x)2-2 × x × 2z+( 2z)2-y2

As we know, x2-2xy+y2=( x-y)2

=  ( x-2z)2-y2

As we know,   p2-q2= ( p + q )( p – q )

= ( x-2z+y) ( x-2z-y)

Therefore, x2-y2-4xz+4z2 =  ( x-2z+y) ( x-2z-y )

 

Q21 . x2+6√2x+10

Solution:

Expanding the middle term ,

= x2+5√2x+√2x+10                                                 [ ∵ 6√2= 5√2 +√2 and 5√2 × √2=10]

= x( x+5√2)+√2( x+5√2)

=  ( x+5√2)( x+√2)

Therefore, x2+6√2x+10= ( x+5√2) ( x+√2)

 

Q22 . x2-2√2 x-30

Solution:

Expanding the middle term,

= x2-5√2 x+3√2 x-30                                                 [ ∵ -2√2 =-5√2 +3√2; also, -5√2 × 3√2 =-30]

= x( x-5√2)+3√2 ( x-5√2)

= ( x-5√2)( x+3√2)

Therefore, x2-2√2 x-30= ( x-5√2)( x+3√2)

 

Q23 . x2-√3x-6

Solution:

Expanding the middle term,

=x2-2√3x+√3x-6                      [ ∵ –√3=-2√3+√3; also, -2√3× √3=-6 ]

= x ( x-2√3)+√3( x-2√3)

=  ( x-2√3)( x+√3)

Therefore, x2√3x-6 = ( x-2√3) ( x+√3)

 

Q24 . x2+5√5x+30

Solution:

Expanding the middle term,

x2+2√5x+3√5x+30                              [ ∵ 5√5=2√5+3√5\;also, 2√5× 3√5=30]

x( x+2√5)+3√5( x+2√5)

= ( x+2√5)(x +3√5 )

Therefore,x2+5√5x+30 = ( x+2√5 )(x +3√5)

 

Q25 . x2+2√3x-24

Solution:

Expanding the middle term,

=x2+4√3x-2√3x-24                                                       [ ∵ 2√3=4√3-2√3;also, 4√3( -2√3)=-24]

= x( x+4√3)-2√3( x+4√3)

= ( x+4√3)( x-2√3)

Therefore, x2+2√3x-24 =  ( x+4√3) ( x-2√3)

 

Q26 . 2x2-5/6 x+1/12                                      

Solution:

Expanding the middle term,

=2x2-x/2-x/3+1/12                          [ ∵ -5/6=-1/2-1/3;also, -1/2 × -1/3 = 2× 1/12]

= x ( 2x-1/2)-1/6 ( 2x-1/2)

= ( 2x-1/2)(x -1/6)

Therefore, 2x2-5/6 x+1/12 = ( 2x-1/2)(x -1/6).

 

Q27 . x2+12/35 x+1/35

Solution:

Expanding the middle term,

=x2+5/35 x+7/35 x+1/35                 [ ∵ 12/35 =5/35 +7/35 and 5/35 × 7/35 =1/35]

= x2+x/7+x/5+1/35

= x ( x+1/7)+1/5( x+1/7)

= ( x+1/7)( x+1/5)

Therefore, x2+12/35 x+1/35 =  ( x+1/7)( x+1/5)

 

Q28 . 21x2-2x+1/21

Solution:

= ( √21x)2-2√21x × 1/√21+( 1/√21)2

As we know, ( x-y)2=x2+y2-2xy

= (√21x-1/√21 )2

Therefore, 21x2-2x+1/21= ( √21x – 1/√21)2

 

Q29 . 5√5x2+20x+3√5

Solution:

Expanding  the middle term,

= 5√5x2+15x+5x+3√5                                           [ ∵ 20=15+5 and 15 × 5=5√5 × 3√5 ]

= 5x( √5x+3 )+√5( √5x+3)

= ( √5x+3)( 5x+√5)

Therefore, 5√5x2+20x+3√5= (√5x+3)( 5x+√5)

 

Q30 .2x2+3√5x+5

Solution:

Expanding the middle term,

= 2x2+2√5x+√5x+5

=  2x ( x+√5)+√5( x+√5)

=  ( x+√5) ( 2x+√5)

Therefore, 2x2+3√5x+5= ( x+√5 )( 2x+√5)

 

Q31 . 9( 2a-b)2-4( 2a-b)-13

Solution:

Say, 2a – b = x

=  9x2-4x-13

Expanding the middle term,

=  9x2-13x+9x-13

=  x( 9x-13)+1( 9x-13)

=  ( 9x-13) ( x+1 )

Substituting  x = 2a – b

=   [ 9 ( 2a-b )-13]( 2a-b+1 )

=  ( 18a-9b -13) ( 2a-b+1 )

Therefore, ( 2a-b)2-4 ( 2a-b)-13 = ( 18a-9b -13)( 2a-b+1)

 

Q 32 . 7( x-2y)2-25 ( x-2y)+12

Solution:

Say,  x-2y = P

=  7P2-25 P+12

Expanding the middle term,

=  7P2-21P-4P+12

=  7P ( P-3)-4( P-3)

=  ( P-3) ( 7P-4)

Substituting,  P = x – 2y, we get;

=   ( x-2y-3) ( 7( x-2y)-4)

= ( x-2y-3) ( 7x-14y-4)

Therefore, 7( x-2y)2-25 ( x-2y)+12= ( x-2y-3)( 7x-14y-4)

 

Q33 . 2( x+y)2– 9 ( x+y )-5

Solution:

Say, x+y = z

=  2z2-9 z-5

Expanding the middle term,

=  2z2-10z+z-5

=  2z ( z-5 )+1 ( z-5 )

=  ( z-5 ) ( 2z+1 )

Substituting,  z = x + y, we get;

=  ( x+y-5 ) ( 2 ( x+y )+1 )

=  ( x+y-5 ) ( 2x+2y+1 )

Therefore, 2( x+y )2-9 ( x+y )-5 = ( x+y-5 ) ( 2x+2y+1 )

 

Q34 . Give the possible expression for the length &  breadth of the rectangle having   35y2-13y-12 as its area.

Solution:

Given, area of rectangle as 35y2– 13y – 12

Expanding the middle term,

Area =  35y2+218y-15y-12

= 7y ( 5y+4 )-3 ( 5y+4 )

( 5y+4) ( 7y-3)

We already know, area of rectangle  = length × breadth

Therefore, It is possible,

length = ( 5y+4) and  breadth= ( 7y-3 )

Or length = ( 7y-3 ) and  breadth= ( 5y+4 )

 

Q35. What are the possible expression for the cuboid having volume 3x– 12x.

Solution:

Volume of the cuboid =  3x2-12x

= 3x ( x-4)

= 3 × x ( x-4)

Also, volume = Length × Breadth × times × Height

Therefore, Possible expression for dimensions of cuboid are = 3 , x , ( x-4)

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