Answer
$-3 \leq x \leq -1~~$ or $~~x \geq 2$
The solution set is $~~[-3, -1]\cup [2,\infty)$
Work Step by Step
$(x+1)(x-2)(x+3) \geq 0$
When $~~x=-1~~$, $~~x = 2,~~$ or $~~x=-3,~~$ then $~~(x+1)(x-2)(x+3) = 0$
When $x \lt -3,~~$ then $~~(x+1)(x-2)(x+3) \lt 0$
When $-3 \lt x \lt -1,~~$ then $~~(x+1)(x-2)(x+3) \gt 0$
When $-1 \lt x \lt 2,~~$ then $~~(x+1)(x-2)(x+3) \lt 0$
When $x \gt 2,~~$ then $~~(x+1)(x-2)(x+3) \gt 0$
Solution:
$-3 \leq x \leq -1~~$ or $~~x \geq 2$
The solution set is $~~[-3, -1]\cup [2,\infty)$
