# RD Sharma Solutions Class 9 Factorization Of Algebraic Expressions Exercise 5.1

## RD Sharma Solutions Class 9 Chapter 5 Ex 5.1

Q1 . $x^{3}+x-3x^{2}-3$

SOLUTION :

Taking  x common in $x^{3}+x$

=$x\left ( x^{2}+1 \right )-3x^{2}-3$

Taking  – 3 common in $-3x^{2}-3$

=$x\left ( x^{2}+1 \right )-3\left ( x^{2}+1 \right )$

Now , we take $\left ( x^{2}+1 \right )$ common

=$\left ( x^{2}+1 \right )$ (x – 3)

$∴$ $x^{3}+x-3y^{2}-3$ =$\left ( x^{2}+1 \right )$ (x – 3)

Q2 . $a\left ( a+b \right )^{3}-3a^{2}b\left ( a+b \right )$

SOLUTION :

Taking  (a + b) common in the two terms

= (a + b) {a(a + b) ² – 3a ²b}

Now,  using  $\left ( a+b \right )^{2}=a^{2}+b^{2}+2ab$

=$\left ( a+b \right )\left \{ a\left ( a^{2}+b^{2}+2ab \right )-3a^{2}b \right \}$

=$\left ( a+b \right )\left \{ a^{3}+ab^{2}+2a^{2}b-3a^{2}b \right \}$

=$\left ( a+b \right )\left \{ a^{3}+ab^{2}-a^{2}b \right \}$

=$\left ( a+b \right )p\left \{ a^{2}+b^{2}-ab \right \}$

=$p\left ( a+b \right )\left ( a^{2}+b^{2}-ab \right )$

$∴ \;a\left ( a+b \right )^{3}-3a^{2}b\left ( a+b \right )=a\left ( a+b \right )\left ( a^{2}+b^{2}-ab \right )$

Q3 . $x\left ( x^{3}-y^{3} \right )+3xy\left ( x-y \right )$

SOLUTION :

Elaborating  $x^{3}-y^{3}$  using the identity  $x^{3}-y^{3}=\left ( x-y \right )\left ( x^{2} +xy+y^{2}\right )$

=$x\left ( x-y \right )\left ( x^{2} +xy+y^{2}\right )+3xy\left ( x-y \right )$

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

=$x\left ( x-y \right )\left ( x^{2}+xy+y^{2}+3y \right )$

$∴$  $x\left ( x^{3}-y^{3} \right )+3xy\left ( x-y \right )$= $x\left ( x-y \right )\left ( x^{2}+xy+y^{2}+3y \right )$

Q4 . $a^{2}x^{2}+\left ( ax^{2}+1 \right )x+a$

SOLUTION :

We multiply $x\left ( ax^{2}+1 \right )=ax^{3}+x$

=$a^{2}x^{2}+ax^{3}+x+a$

Taking common $ax^{2}$ in $\left ( a^{2}x^{2}+ax^{3} \right )$ and  1in ( x + a )

=$ax^{2}\left ( a+x \right )+1\left ( x+a \right )$

=$ax^{2}\left ( a+x \right )+1\left ( a+x \right )$

Taking ( a + x ) common in both the terms

=$\left ( a+x \right )\left ( ax^{2}+1 \right )$

$∴$ $a^{2}x^{2}+\left ( ax^{2}+1 \right )x+a$ =$\left ( a+x \right )\left ( ax^{2}+1 \right )$

Q5 . $x^{2}+y-xy-x$

SOLUTION :

On rearranging

$x^{2}-xy-x+y$

Taking x common in the  $\left ( x^{2}-xy \right )$ and -1 in(-x+y )

=x( x – y ) – 1 ( x – y )

Taking ( x – y ) common in the terms

=( x – y )( x – 1 )

latex]∴\)$x^{2}+y-xy-x$ =( x – y )( x – 1 )

Q6 . $x^{3}-2x^{2}b+3xy^{2}-6y^{3}$

SOLUTION :

Taking $x^{2}$ common in $\left ( x^{3}-2x^{2}y \right )$ and +$3y^{2}$ common in  $\left ( 3xy^{2}-6y^{3} \right )$

= $x^{2}\left ( x-2y \right )+3y^{2}\left ( x-2y \right )$

Taking  ( x – 2y ) common in the terms

= $\left ( x-2y \right )\left ( x^{2}+3y^{2} \right )$

latex]∴\) $x^{3}-2x^{2}y+3xy^{2}-6y^{3}$ = $\left ( x-2y \right )\left ( x^{2}+3y^{2} \right )$

Q7 . $6ab-b^{2}+12ac-2bc$

SOLUTION :

Taking  b common in $\left ( 6ab-b^{2} \right )$ and 2c in ( 12ac – 2bc )

=b( 6a – b ) + 2c ( 6a – b )

Taking  ( 6a – b ) common in the terms

=( 6a – b )( b + 2c )

latex]∴\)$6ab-b^{2}+12ac-2bc$ =( 6a – b )( b + 2c )

Q8 . $\left [ x^{2}+\frac{1}{x^{2}} \right ]-4\left [ x+\frac{1}{x} \right ]+6$

SOLUTION :

=$x^{2}+\frac{1}{x^{2}}-4x-\frac{4}{x}+4+2$

=$x^{2}+\frac{1}{x^{2}}+4+2-\frac{4}{x}-4x$

=$\left ( x^{2} \right )+\left ( \frac{1}{x} \right )^{2}+\left ( -2 \right )^{2}+2\times x\times \frac{1}{x}+2\times \frac{1}{x}\times \left ( -2 \right )+2\left ( -2 \right )x$

Using identity

$x^{2}+y^{2}+z^{2}+2xy+2yz+2zx=\left ( x+y+z \right )^{2}$

We get,

=$\left [ x+\frac{1}{x} +\left ( -2 \right )\right ]^{2}$

=$\left [ x+\frac{1}{x} -2\right ]^{2}$

=$\left [ x+\frac{1}{x} -2\right ]\left [ x+\frac{1}{x} -2\right ]$

$∴$ $\left [ x^{2}+\frac{1}{x^{2}} \right ]-4\left [ x+\frac{1}{x} \right ]+6$ =$\left [ x+\frac{1}{x} -2\right ]\left [ x+\frac{1}{x} -2\right ]$

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 ) $\left \{ x^{2}-4x+4 \right \}$

Now splitting the middle term of $x^{2}-4x+4$

=( x – 2 ){ $x^{2}-2x-2x+4$}

=( x – 2 ){ x( x – 2 ) -2( x -2 )}

=( x – 2 ){( x – 2 )( x- 2 )}

=( x – 2 )( x – 2 )( x – 2 )

=$\left ( x-2 \right )^{3}$

$∴$ x( x- 2 )( x – 4 ) + 4x – 8=$\left ( x-2 \right )^{3}$

Q10 .( x + 2 ) $\left ( x^{2}+25 \right )-10x^{2}-20x$

SOLUTION :

( x + 2 ) $\left ( x^{2}+25 \right )$-10x ( x + 2 )

Taking  ( x + 2 ) common in both the terms

=( x + 2 ) $\left ( x^{2}+25-10x \right )$

=( x + 2 ) $\left ( x^{2}-10x+25 \right )$

Splitting the middle term of  $\left ( x^{2}-10x+25 \right )$

=( x + 2 ) $\left ( x^{2}-5x-5x+25 \right )$

=( x + 2 ){ x (x – 5 ) -5 ( x – 5 )}

=( x + 2 )( x – 5 )( x – 5 )

$∴$ ( x + 2 ) $\left ( x^{2}+25 \right )-10x^{2}$ – 20x=( x + 2 )( x – 5 )( x – 5 )

Q11 . $2a^{2}+2\sqrt{6}ab+3b^{2}$

SOLUTION :

= $\left ( \sqrt{2}a \right )^{2}+2\times \sqrt{2}a\times \sqrt{3}b+\left ( \sqrt{3}b \right )^{2}$

Using the identity $\left ( p+q \right )^{2}=p^{2}+q^{2}+2pq$

= $\left ( \sqrt{2} a+\sqrt{3}b\right )^{2}$

= $\left ( \sqrt{2} a+\sqrt{3}b\right )\left ( \sqrt{2} a+\sqrt{3}b\right )$

$∴$ $2a^{2}+2\sqrt{6}ab+3b^{2}$ = $\left ( \sqrt{2} a+\sqrt{3}b\right )\left ( \sqrt{2} a+\sqrt{3}b\right )$

Q 12 . $(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c)\times (b-c+a)$

SOLUTION :

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

• $=x^{2}+y^{2}+2xy$

Using  the  identity  $\left ( a+b \right )^{2}=a^{2}+b^{2}+2ab$

$=(x+y)^{2}$

Now , substituting  x and y

• $(a–b+c+b-c+a)^{2}$

Cancelling –b , +b  &  +c , -c

$=(2a)^{2}$

$=4a^{2}$

$∴ (a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c)\times (b-c+a) =4a^{2}$

Q13 . $a^{2}+b^{2}+2\left ( ab+bc+ca \right )$

SOLUTION :

$=a^{2}+b^{2}+2ab+2bc+2ca$

Using  the  identity  $\left ( p+q \right )^{2}=p^{2}+q^{2}+2pq$

We get,

= $\left ( a+b \right )^{2}$ + 2bc + 2ca

= $\left ( a+b \right )^{2}$ + 2c( b + a )

Or $\left ( a+b \right )^{2}$ + 2c( a + b )

Taking  ( a + b ) common

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

$∴$ $a^{2}+b^{2}+2\left ( ab+bc+ca \right )$ = ( 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

= $4x^{2}-12xy+9y^{2}$

Splitting the middle term  – 12 = -6 -6  also  $4\times 9=-6\times -6$

=$4x^{2}-6xy-6xy+9y^{2}$

=2x( 2x – 3y ) -3y( 2x – 3y )

=( 2x – 3y ) ( 2x – 3y )

=$\left ( 2x-3y \right )^{2}$

Substituting  x =  x – y  & y =  x + y

=$\left [ 2\left ( x-y \right )-3\left ( x+y \right ) \right ]^{2}$=[ 2x – 2y – 3x – 3y ]2

=(2x-3x-2y-3y )²

=$\left [ -x-5y \right ]^{2}$

=$\left [ \left ( -1 \right )\left ( x+5y \right ) \right ]^{2}$

=$\left ( x+5y \right )^{2}$                                                   [$∵$ (-1)2 = 1]

$∴$ $4(x-y)^{2}-12(x-y)(x+y)+9(x+y)^{2}$=$\left ( x+5y \right )^{2}$

Q 15 . $a^{2}-b^{2}+2bc-c^{2}$

SOLUTION :

$a^{2}-\left ( b^{2}-2bc+c^{2} \right )$

Using the identity  $\left ( a-b \right )^{2}=a^{2}+b^{2}-2ab$

= $a^{2}-\left ( b-c \right )^{2}$

Using the identity  $a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )$

=( a + b – c )( a – ( b – c ))

=( a + b – c )( a – b + c )

$∴$ $a^{2}-b^{2}+2bc-c^{2}$=( a + b – c )( a – b + c )

Q16 . $a^{2}+2ab+b^{2}-c^{2}$

SOLUTION :

Using  the  identity  $\left ( p+q \right )^{2}=p^{2}+q^{2}+2pq$

=$\left ( a+b \right )^{2}-c^{2}$

Using the identity  $p^{2}-q^{2}=\left ( p+q \right )\left ( p-q \right )$

=( a + b + c )( a + b – c )

$∴$ $a^{2}+2ab+b^{2}-c^{2}$ =( a + b + c )( a + b – c )

Q 17 . $a^{2}+4b^{2}- 4ab-4c^{2}$

SOLUTION :

On rearranging

= $a^{2}- 4ab+4b^{2}-4c^{2}$

= $\left ( a \right )^{2}-2\times a\times 2b+\left ( 2b \right )^{2}-4c^{2}$

Using the identity  $\left ( a-b \right )^{2}=a^{2}+b^{2}-2ab$

=$\left ( a-2b \right )^{2}-4c^{2}$

=$\left ( a-2b \right )^{2}-\left ( 2c \right )^{2}$

Using the identity  $a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )$

=( a – 2b  – 2c) ( a – 2b  + 2c)

$∴$ $a^{2}+4b^{2}- 4ab-4c^{2}$ =( a – 2b  – 2c) ( a – 2b  + 2c)

Q18 . $xy^{9}-yx^{9}$

SOLUTION :

=$xy\left ( y^{8}-x^{8} \right )$

=$xy\left ( \left ( y^{4} \right )^{2}-\left ( x^{4} \right )^{2} \right )$

Using the identity  $p^{2}-q^{2}$= ( p + q )( p – q )

=$xy\left ( y^{4}+x^{4} \right )\left ( y^{4}-x^{4} \right )$

=$xy\left ( y^{4}+x^{4} \right )\left ( \left ( y^{2} \right )^{2}-\left ( x^{2} \right )^{2} \right )$

Using the identity  $p^{2}-q^{2}$= ( p + q )( p – q )

=$xy\left ( y^{4}+x^{4} \right )\left ( y^{2}+x^{2} \right )\left ( y^{2}-x^{2} \right )$

= $xy\left ( y^{4}+x^{4} \right )\left ( y^{2}+x^{2} \right )\left ( y+x \right )\left ( y-x \right )$

= $xy\left ( x^{4}+y^{4} \right )\left ( x^{2}+y^{2} \right )\left ( x+y \right )\left ( -1 \right )\left ( x-y \right )$

$∵ \left ( y-x \right )=-1\left ( x-y \right )$

=$-xy\left ( x^{4}+y^{4} \right )\left ( x^{2}+y^{2} \right )\left ( x+y \right )\left ( x-y \right )$

$∴$ $xy^{9}-yx^{9}$ = $-xy\left ( x^{4}+y^{4} \right )\left ( x^{2}+y^{2} \right )\left ( x+y \right )\left ( x-y \right )$

Q 19 . $x^{4}+x^{2}y^{2}+y^{4}$

SOLUTION :

Adding  $x^{2}y^{2}$ and subtracting $x^{2}y^{2}$ to the given equation

= $x^{4}+x^{2}y^{2}+y^{4}+x^{2}y^{2}-x^{2}y^{2}$

= $x^{4}+2x^{2}y^{2}+y^{4}-x^{2}y^{2}$

= $\left ( x^{2} \right )^{2}+2\times x^{2}\times y^{2}+\left ( y^{2} \right )^{2}-\left ( xy \right )^{2}$

Using  the  identity  $\left ( p+q \right )^{2}=p^{2}+q^{2}+2pq$

= $\left ( x^{2}+y^{2} \right )^{2}-\left ( xy \right )^{2}$

Using the identity  $p^{2}-q^{2}$= ( p + q )( p – q )

= $\left ( x^{2}+y^{2}+xy \right )\left ( x^{2}+y^{2}-xy\right )$

$∴$ $x^{4}+x^{2}y^{2}+y^{4}$ = $\left ( x^{2}+y^{2}+xy \right )\left ( x^{2}+y^{2}-xy\right )$

Q20 . $x^{2}-y^{2}-4xz+4z^{2}$

SOLUTION :

On rearranging the terms

= $x^{2}-4xz+4z^{2}-y^{2}$

= $\left ( x \right )^{2}-2\times x\times 2z+\left ( 2z \right )^{2}-y^{2}$

Using the identity $x^{2}-2xy+y^{2}=\left ( x-y \right )^{2}$

= $\left ( x-2z \right )^{2}-y^{2}$

Using the identity  $p^{2}-q^{2}$= ( p + q )( p – q )

= $\left ( x-2z+y \right )\left ( x-2z-y \right )$

$∴$ $x^{2}-y^{2}-4xz+4z^{2}$ = $\left ( x-2z+y \right )\left ( x-2z-y \right )$

Q21 . $x^{2}+6\sqrt{2}x+10$

SOLUTION :

Splitting the middle term ,

= $x^{2}+5\sqrt{2}x+\sqrt{2}x+10$                                                  $\left [ ∵ 6\sqrt{2}= 5\sqrt{2} +\sqrt{2}\; and\; 5\sqrt{2}\times \sqrt{2}=10 \right ]$

= $x\left ( x+5\sqrt{2} \right )+\sqrt{2}\left ( x+5\sqrt{2} \right )$

= $\left ( x+5\sqrt{2} \right )\left ( x+\sqrt{2} \right )$

$∴$ $x^{2}+6\sqrt{2}x+10$ = $\left ( x+5\sqrt{2} \right )\left ( x+\sqrt{2} \right )$

Q22 . $x^{2}-2\sqrt{2}x-30$

SOLUTION :

Splitting the middle term,

= $x^{2}-5\sqrt{2}x+3\sqrt{2}x-30$                                                  $\left [ ∵ -2\sqrt{2} =-5\sqrt{2}+3\sqrt{2}\;also\;-5\sqrt{2}\times 3\sqrt{2}=-30\right ]$

= $x\left ( x-5\sqrt{2} \right )+3\sqrt{2}\left ( x-5\sqrt{2} \right )$

= $\left ( x-5\sqrt{2} \right )\left ( x+3\sqrt{2} \right )$

$∴$ $x^{2}-2\sqrt{2}x-30$ = $\left ( x-5\sqrt{2} \right )\left ( x+3\sqrt{2} \right )$

Q23 . $x^{2}-\sqrt{3}x-6$

SOLUTION :

Splitting the middle term,

=$x^{2}-2\sqrt{3}x+\sqrt{3}x-6$                       $\left [ ∵ -\sqrt{3}=-2\sqrt{3}+\sqrt{3}\;also\;-2\sqrt{3 }\times \sqrt{3}=-6 \right ]$

= $x\left ( x-2\sqrt{3} \right )+\sqrt{3}\left ( x-2\sqrt{3} \right )$

= $\left ( x-2\sqrt{3} \right )\left ( x+\sqrt{3} \right )$

$∴$ $x^{2}-\sqrt{3}x-6$ = $\left ( x-2\sqrt{3} \right )\left ( x+\sqrt{3} \right )$

Q24 . $x^{2}+5\sqrt{5}x+30$

SOLUTION :

Splitting the middle term,

= $x^{2}+2\sqrt{5}x+3\sqrt{5}x+30$                                $\left [ ∵ 5\sqrt{5}=2\sqrt{5}+3\sqrt{5}\;also\;2\sqrt{5}\times 3\sqrt{5}=30 \right ]$

= $x\left ( x+2\sqrt{5} \right )+3\sqrt{5}\left ( x+2\sqrt{5} \right )$

= $\left ( x+2\sqrt{5} \right )\left (x +3\sqrt{5} \right )$

$∴$ $x^{2}+5\sqrt{5}x+30$ = $\left ( x+2\sqrt{5} \right )\left (x +3\sqrt{5} \right )$

Q25 . $x^{2}+2\sqrt{3}x-24$

SOLUTION :

Splitting the middle term,

=$x^{2}+4\sqrt{3}x-2\sqrt{3}x-24$                                   $\left [ ∵ 2\sqrt{3}=4\sqrt{3}-2\sqrt{3}\;also\;4\sqrt{3}\left ( -2\sqrt{3} \right )=-24 \right ]$

= $x\left ( x+4\sqrt{3} \right )-2\sqrt{3}\left ( x+4\sqrt{3} \right )$

= $\left ( x+4\sqrt{3} \right )\left ( x-2\sqrt{3} \right )$

$∴$ $x^{2}+2\sqrt{3}x-24$ = $\left ( x+4\sqrt{3} \right )\left ( x-2\sqrt{3} \right )$

Q26 . $2x^{2}-\frac{5}{6}x+\frac{1}{12}$

SOLUTION :

Splitting the middle term,

= $2x^{2}-\frac{x}{2}-\frac{x}{3}+\frac{1}{12}$                           $\left [ ∵ -\frac{5}{6}=-\frac{1}{2}-\frac{1}{3}\;also\;-\frac{1}{2}\times -\frac{1}{3}=2\times \frac{1}{12} \right ]$

= $x\left ( 2x-\frac{1}{2} \right )-\frac{1}{6}\left ( 2x-\frac{1}{2} \right )$

= $\left ( 2x-\frac{1}{2} \right )\left (x -\frac{1}{6} \right )$

$∴$ $2x^{2}-\frac{5}{6}x+\frac{1}{12}$ = $\left ( 2x-\frac{1}{2} \right )\left (x -\frac{1}{6} \right )$

Q27 . $x^{2}+\frac{12}{35}x+\frac{1}{35}$

SOLUTION :

Splitting the middle term,

=$x^{2}+\frac{5}{35}x+\frac{7}{35}x+\frac{1}{35}$                   $\left [ ∵ \frac{12}{35}=\frac{5}{35}+\frac{7}{35} \;and\;\frac{5}{35}\times \frac{7}{35}=\frac{1}{35}\right ]$

= $x^{2}+\frac{x}{7}+\frac{x}{5}+\frac{1}{35}$

= $x\left ( x+\frac{1}{7} \right )+\frac{1}{5}\left ( x+\frac{1}{7} \right )$

= $\left ( x+\frac{1}{7} \right )\left ( x+\frac{1}{5} \right )$

$∴$ $x^{2}+\frac{12}{35}x+\frac{1}{35}$ = $\left ( x+\frac{1}{7} \right )\left ( x+\frac{1}{5} \right )$

Q28 . $21x^{2}-2x+\frac{1}{21}$

SOLUTION :

= $\left ( \sqrt{21x} \right )^{2}-2\sqrt{21}x\times \frac{1}{\sqrt{21}}+\left ( \frac{1}{\sqrt{21}} \right )^{2}$

Using the identity  $\left ( x-y \right )^{2}=x^{2}+y^{2}-2xy$

= $\left ( \sqrt{21}x-\frac{1}{\sqrt{21}} \right )^{2}$

$∴$ $21x^{2}-2x+\frac{1}{21}$ = $\left ( \sqrt{21}x-\frac{1}{\sqrt{21}} \right )^{2}$

Q29 . $5\sqrt{5}x^{2}+20x+3\sqrt{5}$

SOLUTION :

Splitting the middle term,

= $5\sqrt{5}x^{2}+15x+5x+3\sqrt{5}$                                             $\left [ ∵ 20=15+5\;and\;15\times 5=5\sqrt{5}\times 3\sqrt{5} \right ]$

= $5x\left ( \sqrt{5}x+3 \right )+\sqrt{5}\left ( \sqrt{5}x+3 \right )$

= $\left ( \sqrt{5}x+3 \right )\left ( 5x+\sqrt{5} \right )$

$∴$ $5\sqrt{5}x^{2}+20x+3\sqrt{5}$ = $\left ( \sqrt{5}x+3 \right )\left ( 5x+\sqrt{5} \right )$

Q30 . $2x^{2}+3\sqrt{5}x+5$

SOLUTION :

Splitting the middle term,

=  $2x^{2}+2\sqrt{5}x+\sqrt{5}x+5$

=  $2x\left ( x+\sqrt{5} \right )+\sqrt{5}\left ( x+\sqrt{5} \right )$

=  $\left ( x+\sqrt{5} \right )\left ( 2x+\sqrt{5} \right )$

$∴$ $2x^{2}+3\sqrt{5}x+5$ = $\left ( x+\sqrt{5} \right )\left ( 2x+\sqrt{5} \right )$

Q31 . $9\left ( 2a-b\right )^{2}-4\left ( 2a-b \right )-13$

SOLUTION :

Let  2a – b = x

=  $9x^{2}-4x-13$

Splitting the middle term,

=  $9x^{2}-13x+9x-13$

=  $x\left ( 9x-13 \right )+1\left ( 9x-13 \right )$

=  $\left ( 9x-13 \right )\left ( x+1 \right )$

Substituting  x = 2a – b

=  $\left [ 9\left ( 2a-b \right )-13 \right ]\left ( 2a-b+1 \right )$

=  $\left ( 18a-9b -13 \right ) \left ( 2a-b+1 \right )$

$∴$ $9\left ( 2a-b\right )^{2}-4\left ( 2a-b \right )-13$ = $\left ( 18a-9b -13 \right ) \left ( 2a-b+1 \right )$

Q 32 . $7\left ( x-2y \right )^{2}-25\left ( x-2y\right )+12$

SOLUTION :

Let  x-2y = P

=  $7P ^{2}-25 P+12$

Splitting the middle term,

=  $7P ^{2}-21P-4P+12$

=  $7P\left ( P-3 \right )-4\left ( P-3 \right )$

=  $\left ( P-3 \right )\left ( 7P-4 \right )$

Substituting  P = x – 2y

=  $\left ( x-2y-3 \right )\left ( 7\left ( x-2y \right )-4 \right )$

=  $\left ( x-2y-3 \right )\left ( 7x-14y-4 \right )$

$∴$ $7\left ( x-2y \right )^{2}-25\left ( x-2y\right )+12$ = $\left ( x-2y-3 \right )\left ( 7x-14y-4 \right )$

Q33 . $2\left ( x+y \right )^{2}-9\left ( x+y \right )-5$

SOLUTION :

Let  x+y = z

=  $2z^{2}-9 z-5$

Splitting the middle term,

=  $2z^{2}-10z+z-5$

=  $2z\left ( z-5 \right )+1\left ( z-5 \right )$

=  $\left ( z-5 \right )\left ( 2z+1 \right )$

Substituting  z = x + y

=  $\left ( x+y-5 \right )\left ( 2\left ( x+y \right )+1 \right )$

=  $\left ( x+y-5 \right )\left ( 2x+2y+1 \right )$

$∴$ $2\left ( x+y \right )^{2}-9\left ( x+y \right )-5$ = $\left ( x+y-5 \right )\left ( 2x+2y+1 \right )$

Q34 . Give the possible expression for the  length  &  breadth of the rectangle having   $35y^{2}-13y-12$ as its area.

SOLUTION :

Area is given as $35y^{2}-13y-12$

Splitting the middle term,

Area =  $35y^{2}+218y-15y-12$

=  $7y\left ( 5y+4 \right )-3\left ( 5y+4 \right )$

$\left ( 5y+4\right )\left ( 7y-3 \right )$

We also know that area of rectangle  = length $\times$breadth

$∴$ Possible length = $\left ( 5y+4 \right )$ and  breadth=$\left ( 7y-3 \right )$

Or  possible length = $\left ( 7y-3 \right )$ and  breadth= $\left ( 5y+4 \right )$

Q35 . What are  the possible expression for the cuboid having volume $3x^{2}-12x$.

SOLUTION :

Volume =  $3x^{2}-12x$

= $3x\left ( x-4 \right )$

= $3\times x\left ( x-4 \right )$

Also volume = Length $\times$Breadth$\times$Height

$∴$ Possible expression for dimensions of cuboid are = 3 , x , $\left ( x-4 \right )$

#### Practise This Question

10sin1(2x1+x2)dx=