Find the solution set for the inequality x2+2x−8x2+x−2≤0\frac{x^2 + 2x - 8}{x^2 + x - 2} \leq 0x2+x−2x2+2x−8≤0.
[−4,−2)∪(1,2][-4, -2) \cup (1, 2][−4,−2)∪(1,2]
(−2,1)∪[2,4](-2, 1) \cup [2, 4](−2,1)∪[2,4]
[−4,−1]∪(1,2][-4, -1] \cup (1, 2][−4,−1]∪(1,2]
(−2,2]∪[4,∞)(-2, 2] \cup [4, \infty)(−2,2]∪[4,∞)