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

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 , \(\left ( x-4 \right )\)