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

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 x²y² to 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)