Question 407321
It doesn't work because if we assume x+1 = 4, we fail to take into account the x-2 expression. However, if we had the equation (x+1)(x-2) = 0, then we can set x+1 = 0, x-2 = 0 because the other term is irrelevant.


The best way is to expand and get {{{x^2 - x - 2 = 4}}} --> {{{x^2 - x - 6 = 0}}}. This factors to {{{(x-3)(x+2) = 0}}} --> x = 3, -2.