RD Sharma Solutions -Ex-5.1, (Part - 1) Factorization Of Algebraic Expressions, Class 9, Maths

# RD Sharma Solutions -Ex-5.1, (Part - 1) Factorization Of Algebraic Expressions, Class 9, Maths - Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

 1 Crore+ students have signed up on EduRev. Have you?

Q1 . x3+x−3x2−3

SOLUTION :

Taking  x common in x3+x

=x(x2+1)−3x2−3

Taking  – 3 common in −3x2−3

=x(x2+1)−3(x2+1)

Now , we take (x2+1) common

=(x2+1) (x – 3)

∴ x3+x−3y2−3 =(x2+1) (x – 3)

Q2 . a(a+b)3−3a2b(a+b)

SOLUTION :

Taking  (a + b) common in the two terms

= (a + b) {a(a + b) ² – 3a ²b}

Now,  using  (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)

∴a(a+b)3−3a2b(a+b)=a(a+b)(a2+b2−ab)

Q3 . x(x3−y3)+3xy(x−y)

SOLUTION :

Elaborating  x3−y3  using the identity  x3−y3=(x−y)(x2+xy+y2)

=x(x−y)(x2+xy+y2)+3xy(x−y)

Taking common x( x-y ) in both the terms

=x(x−y)(x2+xy+y2+3y)

∴  x(x3−y3)+3xy(x−y)= x(x−y)(x2+xy+y2+3y)

Q4 . a2x2+(ax2+1)x+a

SOLUTION :

We multiply x(ax2+1)=ax3+x

=a2x2+ax3+x+a

Taking common ax2 in (a2x2+ax3) and  1in ( x + a )

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

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

Taking ( a + x ) common in both the terms

=(a+x)(ax2+1)

∴ a2x2+(ax2+1)x+a =(a+x)(ax2+1)

Q5 . x2+y−xy−x

SOLUTION :

On rearranging

x2−xy−x+y

Taking x common in the  (x2−xy) and -1 in(-x+y )

=x( x – y ) – 1 ( x – y )

Taking ( x – y ) common in the terms

=( x – y )( x – 1 )

latex]∴\)x2+y−xy−x =( x – y )( x – 1 )

Q6 . x3−2x2b+3xy2−6y3

SOLUTION :

Taking x2 common in (x3−2x2y) and +3y2 common in  (3xy2−6y3)

= x2(x−2y)+3y2(x−2y)

Taking  ( x – 2y ) common in the terms

= (x−2y)(x2+3y2)

latex]∴\) x3−2x2y+3xy2−6y3 = (x−2y)(x2+3y2)

Q7 . 6ab−b2+12ac−2bc

SOLUTION :

Taking  b common in (6ab−b2) and 2c in ( 12ac – 2bc )

=b( 6a – b ) + 2c ( 6a – b )

Taking  ( 6a – b ) common in the terms

=( 6a – b )( b + 2c )

latex]∴\)6ab−b2+12ac−2bc =( 6a – b )( b + 2c )

Q8 .

SOLUTION :

Using identity

x2+y2+z2+2xy+2yz+2zx=(x+y+z)2

We get,

Q9 . x( x- 2 )( x – 4 ) + 4x – 8

SOLUTION :

=x( x  – 2 )( x – 4 ) + 4( x – 2 )

Taking ( x – 2 ) common in both the terms

=( x – 2 ){x( x – 4 ) + 4}

=( x – 2 ) {x2−4x+4}

Now splitting 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

∴ 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 )

Taking  ( x + 2 ) common in both the terms

=( x + 2 ) (x2+25−10x)

=( x + 2 ) (x2−10x+25)

Splitting 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 ) (x2+25)−10x2 – 20x=( x + 2 )( x – 5 )( x – 5 )

Q11 .

SOLUTION :

Using the identity (p+q)2=p2+q2+2pq

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

SOLUTION :

Let  ( a – b + c ) = x and ( b – c + a ) = y

• =x2+y2+2xy

Using  the  identity  (a+b)2=a2+b2+2ab

=(x+y)2

Now , substituting  x and y

• (a–b+c+b−c+a)2

Cancelling –b , +b  &  +c , -c

=(2a)2

=4a2

∴(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

Using  the  identity  (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 )

Taking  ( a + b ) common

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

∴ 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 :

Let ( x – y ) = x,( x + y ) = y

= 4x2−12xy+9y2

Splitting 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]

∴ 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)

Using the identity  (a−b)2=a2+b2−2ab

= a2−(b−c)2

Using the identity  a2−b2=(a+b)(a−b)

=( a + b – c )( a – ( b – c ))

=( a + b – c )( a – b + c )

∴ a2−b2+2bc−c2=( a + b – c )( a – b + c )

Q16 . a2+2ab+b2−c2

SOLUTION :

Using  the  identity  (p+q)2=p2+q2+2pq

=(a+b)2−c2

Using the identity  p2−q2=(p+q)(p−q)

=( a + b + c )( a + b – c )

∴ 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

Using the identity  (a−b)2=a2+b2−2ab

=(a−2b)2−4c2

=(a−2b)2−(2c)2

Using the identity  a2−b2=(a+b)(a−b)

=( a – 2b  – 2c) ( a – 2b  + 2c)

∴ a2+4b2−4ab−4c2 =( a – 2b  – 2c) ( a – 2b  + 2c)

Q18 . xy9−yx9

SOLUTION :

=xy(y8−x8)

=xy((y4)2−(x4)2)

Using the identity  p2−q2= ( p + q )( p – q )

=xy(y4+x4)(y4−x4)

=xy(y4+x4)((y2)2−(x2)2)

Using the identity  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)

∴ xy9−yx9 = −xy(x4+y4)(x2+y2)(x+y)(x−y)

Q 19 . x4+x2y2+y4

SOLUTION :

Adding  x2y2 and subtracting x2yto the given equation

= x4+x2y2+y4+x2y2−x2y2

= x4+2x2y2+y4−x2y2

= (x2)2+2×x2×y2+(y2)2−(xy)2

Using  the  identity  (p+q)2=p2+q2+2pq

= (x2+y2)2−(xy)2

Using the identity  p2−q2= ( p + q )( p – q )

= (x2+y2+xy)(x2+y2−xy)

∴ x4+x2y2+y4 = (x2+y2+xy)(x2+y2−xy)

The document RD Sharma Solutions -Ex-5.1, (Part - 1) Factorization Of Algebraic Expressions, Class 9, Maths - Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9 is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
All you need of Class 9 at this link: Class 9

## RD Sharma Solutions for Class 9 Mathematics

91 docs
 Use Code STAYHOME200 and get INR 200 additional OFF

## RD Sharma Solutions for Class 9 Mathematics

91 docs

### Top Courses for Class 9

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;