(x+1)(x-2)>(x+1)(-x+6) take the rhs over to the lhs
(x+1)(x-2) - (x+1)(-x+6) > 0 take out the common factor of (x+1)
(x+1)(x-2 +x-6) > 0
(x+1)(2x-8) > 0 divide by 2
(x+1)(x-4) > 0
For the above inequality to be true, we require (x+1) and (x-4) to both be positive OR (x+1) and (x-4) to both be negative. i.e.
(x+1) > 0 and (x-4) > 0 OR (x+1) < 0 and (x-4) < 0
x > -1 and x > 4 OR x < -1 and x < 4
Now the first option, x > -1 and x > 4, is satisfied by x > 4.
The second option, x < -1 and x < 4, is satisfied by x < -1.
We thus require x to be constrained by
x < -1 or x > 4