Please solve those questions seperately for me

Answer :

We have  ( x +  2 ) ( x +  3 ) + ( x - 3 ) ( x -  2 ) - 2x ( x +  1 ) = 0

So,

$\Rightarrow$( x² +  3x  + 2x  + 6 )  +( x² -  3x  - 2x  + 6 ) - ( 2x² + 2x ) = 0

$\Rightarrow$( x² +  5x  + 6 )  +( x² -  5x  + 6 ) - ( 2x² + 2x ) = 0

 $\Rightarrow$x² +  5x  + 6  +  x² -  5x  + 6 -  2x² - 2x  = 0

$\Rightarrow$12  - 2x  =  0

$\Rightarrow$2x  =  12

$\Rightarrow$x  =  6  ( Ans )