**Question:**

(X+1)(X+2)(X+3)(X+4)-8=0 FIND THE VALUE OF 'X'?????????

+2)(x+3)-8=0 , (x^2+5x+4)(x^2+5x

+6)-8=0 , let x^2+5x=t , (t+4)(t+6)-8=0 ,

t^2+10t+16=0 ,(t+8)(t+2)=0 , t=-8 and

t=-2 , x^2+5x+8=0 and x^2+5x+2=0 , x=

(-5+√7i)/2 ,(-5-√7i)/2 and x=(-5+√17i)/2 , (-5-√17i)/2

7, 6, 5, 4

If you pass the number eight to the other side you will have each one of the solutions