Answer
$\text{the interval } [-2,3]$
Work Step by Step
Equating each factor to zero results to
\begin{array}{l}
x+2=0\\
x=-2
,\\\\\text{OR}\\\\
x-3=0\\
x=3
.\end{array}
Hence, the critical points are $x=\{-2,3\}$.
If $x\lt-2$, then,
\begin{array}{l}
\dfrac{x+2}{x-3}
\\\Rightarrow
\dfrac{(-)}{(-)}
\\=
\text{ positive}
.\end{array}
If $-2\lt x\lt3$, then,
\begin{array}{l}
\dfrac{x+2}{x-3}
\\\Rightarrow
\dfrac{(+)}{(-)}
\\=
\text{ negative}
.\end{array}
If $x\gt3$, then,
\begin{array}{l}
\dfrac{x+2}{x-3}
\\\Rightarrow
\dfrac{(+)}{(+)}
\\=
\text{ positive}
.\end{array}
Since the given inequality uses $\le0$, then take the negative result. Hence, the solution set is $
\text{the interval } [-2,3].
$
