#### Answer

-1 and 2
[-1,2)

#### Work Step by Step

The zeroes are
(x-2)— > x=2
(x+1)— > x=-1
In this case the inquality has to be less than or equal to zero, therefore we know that x=2 can’t work because a 0 in the denominator will make it undefined.
In this case the real line will be divided into:
$(-\infty, -1] [-1, 2) (2, \infty)$
After that you have to choose one value between the value and test it
— — — — — — — — - $(-\infty, -1]— — — — [-1, 2)— —— (2, \infty)$
Sign of x+1— — — (-2+1)= -1 (-)— — — — (1+1)=2 (+)— —(3+1)=4 (+)
Sign of x-2 — — —(-2-2)=-4 (-)— — — — (1-2)=-1 (-)— — (3-2)=1 (+)
(x+1)(x+2)— — — — —(+)— — — — — — — — -(—)— — — — — — — (+)
The intervals that satisfy the inequality are [-1,2)