RD Sharma Solutions Class 9 Factorization Of Algebraic Expressions Exercise 5.1

RD Sharma Solutions Class 9 Chapter 5 Ex 5.1

Q1 . x³ + x – 3x² – 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 x² as common factor from  ( x³-2x²y) and + 3y² common 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 :

=x² + 1/x² – 4x – 4/x + 4 + 2

=x² + 1/x² + 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 × x² × 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 3x² – 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)