How can I solve $\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0$ and express solution in interval notation? The inequality I need to solve is
$$\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0$$
My attempt:
Case I: $x<2$
Then we have $(1-x)(x+3)\leq0$
$x\leq-3, x\geq1$
We reject $x\leq-3$ (not sure why exactly, I just looked at the graph on desmos and it shows that $x\geq1$
Case II: $x<-1$
Then we have $(1-x)(x+3)\geq0$
$x\geq-3, x\leq-1$
Reject $x\leq-1$ (again, not sure exactly why aside from the graph I saw on desmos)
So the final solution is $x\in[-3,-1) \cup [1,2)$
I know that my final solution is correct, however I am a little confused on why I reject certain values.

Case I: $\ (1-x)(x+3) \le 0$ and $(x+1)(2-x) > 0$
\begin{align}
&(1-x)(x+3) \le 0 &\text{AND} \qquad&(x+1)(2-x) > 0\\
&\Rightarrow\begin{cases}
x \ge 1\ \&\ x \ge -3 \text{ OR}\\
x \le 1\ \&\ x \le -3
\end{cases}&\text{AND} \qquad
&\Rightarrow\begin{cases}
x > -1\ \&\ x < 2 \text{ OR}\\
x < -1\ \&\ x > 2 \text{ (Impossible)}\\
\end{cases} \\
&\Rightarrow\begin{cases}
x \ge 1 \text{ OR}\\
x \le -3
\end{cases}&\text{AND} \qquad
&\Rightarrow
-1 < x < 2
\end{align}
$\therefore\ 1 \le x < 2$
Case II: $\ (1-x)(x+3) \ge 0$ and $(x+1)(2-x) < 0$
\begin{align}
&(1-x)(x+3) \ge 0 &\text{AND} \qquad&(x+1)(2-x) < 0\\
&\Rightarrow\begin{cases}
x \le 1 \ \&\ x \ge -3 \text{ OR}\\
x \ge 1 \ \&\ x \le -3 \text{ (Impossible)}
\end{cases}&\text{AND} \qquad
&\Rightarrow\begin{cases}
x < -1 \ \&\ x < 2 \text{ OR}\\
x > -1 \ \&\ x > 2 \\
\end{cases} \\
&\Rightarrow -3 \le x \le 1
&\text{AND} \qquad
&\Rightarrow\begin{cases}
x < -1 \text{ OR}\\
x > 2
\end{cases} \\
\end{align}
$\therefore\ -3 \le x < -1$
Hence, $x\in[-3,-1)\ \cup\ [1, 2)$.
