Question 32447
x^2 + 2xy + 2x + y^2 + 2y - 8

there are different ways through which you can factorise the equation...I am here factorising upto least level/...

here we edit this problem and we get

=(x^2 + y^2 + 2xy) + (2x + 2y) - 8

Now see carefully, you find that in first three variables in brackets you can apply the following formula.
(a+b)^2 = (a)^2 + (b)^2 + 2(a)(b).....
here a = x and b = y ...

on applying the formula we get..

= (x+y)^2 +  (2x + 2y) - 8

now take 2 commom from (2x + 2y) and we get

= (x+y)^2 +  2(x + y) - 8

Now take (x+y)common from (x+y)^2 + 2(x + y) and we get....

= (x+y) [(x+y) - 2 ] - 8 
 
on solving brackets we get...

= (x+y) [x+y - 2 ] - 8 

= (x+y) x + y - 2 - 8 

= (x+y) (x + y - 6)

Hope this will help you 

Please feel free to revert back for any further queries