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.
 A: We have that
$$I(x)=\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0 \iff I(x)=\frac{(x-1)(x+3)}{(x+1)(x-2)}\leq0$$
then consider the intervals intercepted by points $x=-3$, $x=-1$, $x=1$ and $x=2$ and then the cases

*

*$x<-3 \implies \frac{(-)\times(-)}{(-)\times(-)}=(+)$


*$-3<x<-1\implies \frac{(-)\times(+)}{(-)\times(-)}=(-)$


*$\cdots$
noting that sign for $I(x)$ alternates
$$ \bbox[yellow]
{
\begin{array}{|c|c|c|c|c|}
\hline
(-\infty,-3)&(-3,-1) & (-1,1) & (1,2)&(2,\infty) \\ \hline
 +&- &+ &-&+\\ \hline
\end{array}
}
$$
and that we need $x \neq -1$ and $x\neq 2$ with $I(x)=0$ for $x=1$ and $x=-3$.
A: Sketch a Graph (by hand)
My preferred method for solving inequalities involving rational functions is to sketch a quick graph.

*

*Examining asymptotic behaviour, we have
$$ \frac{(1-x)(x+3)}{(x+1)(2-x)} = \frac{-x^2 + \text{junk}}{-x^2 + \text{junk}} \sim 1,$$
hence the graph has a horizontal asymptote at $y=1$.  Note:  if you don't like "junk" and want something a little more formal, you can replace it with $O(x)$.


*The function on the left has two roots of multiplicity one at $x=-3$ and at $x=1$ (multiplicity one means that it crosses the $x$-axis at those roots), and two poles of order one at $x=-1$ and $x=2$ (thus the graph has vertical asymptotes at these points; in both cases, the order is odd, so there is a sign change across the asymptote).
Combining these data, I get the following graph:

The original inequality is satisfied whenever the graph is at or below the $x$-axis, thus
$$ \frac{(1-x)(x+3)}{(x+1)(2-x)} \le 0 \iff x \in [-3,-1) \cup [1,2). $$
I find this approach satisfying four a couple of reasons:

*

*It took me longer to type up everything written above than it did to solve the problem, by at least an order of magnitude.  Indeed, if a problem like this is meant to be solved by hand, then graphing is often very quick (there might be a short factoring step involved, but even then, it should go fast).


*There is almost zero computation, which means that there is very little chance that I will make a mistake.  Moreover, there is a moderate error check in the procedure, in that I can easily catch certain sign errors (I know what the asymptotic behaviour is supposed to look like, and I know how many times the graph crosses the $x$-axis, so I can quickly correct an odd number of sign errors while graphing).

Construct a Sign Table
Another approach, alluded to by several of the other answers, is to think carefully about the signs of the factors.  First, observe that the left-hand side is zero when $x = 1$ and when $x = -3$, thus we immediately obtain two solutions.  Moreover, the left-hand side is undefined when $x=-1$ and when $x=2$ (that is, we cannot have either $x=-1$ or $x=2$).
These key points divide the rest of the real line into five open intervals:
$$ (-\infty,-3),\quad (-3,-1),\quad (-1,1),\quad (1,2),\quad (2,\infty).$$
The function is the product of several factors, and the sign of a product is determined by the parity of the number of negative factors.  So we can look at the sign of each factor on each of the intervals above, then consider the parity of negative factors.  This is summarized in the following "sign table":
$$
\begin{array}{|c|ccccc|}
\hline
& (-\infty, -3) & (-3,-1) & (-1,1) & (1,2) & (2,\infty) \\\hline
1-x & + & + & + & - & - \\
x+3 & - & + & + & + & + \\
x+1 & - & - & + & + & - \\
2-x & + & + & + & + & - \\\hline
\text{Parity} & \text{even} & \text{odd} & \text{even} & \text{odd} & \text{even} \\
\text{Sign} & + & - & + & - & + \\\hline
\end{array}
$$
Therefore the inequality holds whenever
$$ x = \{1\} \cup \{-3\} \cup (-3,-1) \cup (1,2)
= [-3,-1) \cup [1,2). $$
I am not really a huge fan of this approach, as it is easy to make silly errors (I made several while typing this up), it is kind of time consuming, and it is easy to forget about the endpoints (that is, I feel like the endpoints have to be handled separately).
However, a lot of my students like this approach, and it isn't wrong, so...
A: $$P(x)=\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0$$
$x=-3$
$x=\color{red}{-1}$
$x=1$
$x=\color{red}2$
are the x-intercept values or vertical asymptotes. After each one, sign of the function will change because of each one occurs only once(odd number time)
And when multiplying sign of each leading coefficient, it will be positive.
By starting with positive sign from right
$$ \bbox
{
\begin{array}{|ccccc|}
\hline
x&-\infty&-3& \color{red}{-1} & 1&\color{red}2&\infty \\ \hline
 P(x)&\hspace{2cm} {+}&\hspace{2cm}- &\hspace{2cm}+ &\hspace{2cm}-&\hspace{2cm}+\\ \hline
\end{array}
}
$$
$-3\leq x < -1$ or $1\leq x<2$
A: Note you have four factors, that are $1-x$, $x+3$, $(x+1)^{-1}$ and $(2-x)^{-1}$. Now the product is negative exactly if an odd number of these factors are negative.
$1-x\leq0$ iff $x\geq 1$.
$x+3\leq0$ iff $x\leq -3$.
$x+1\leq0$ iff $x\leq -1$.
$2-x\leq0$ iff $x\geq 2$.
So if $x< -3$ two factor is $< 0$, so it’s postive. For $-3<x<-1$ one factor is negative, so it’s negative. For $-1<x<1$ we have no factor negative, so the product is positive. For $1<x<2$ we have one factor negative, so negative and for $x>2$ two factors are negative, so postive.
When is it $0$? Only if $x=1$ or $x=-3$. So we get
$[-3,-1)\cup[1,2)$
EDIT: By the way, you can verify this by
https://sagecell.sagemath.org/?z=eJwryMkv0dAw1K3Q1NKo0DbW1NcAUoZAjhFQSFNH11THVKcyN7HC1hhIZebZ6hprAgBXBg1Q&lang=sage&interacts=eJyLjgUAARUAuQ==
A: 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, $x\in[-3,-1)\ \cup\ [1, 2)$.
A: The key insight here is that for a fraction to be negative you need to have the numerator and denominator have opposite signs.  If you have $$\frac{A}{B}\le 0$$ then either you have $A\le 0$ and $B> 0$ or you have $A\ge 0$ and $B< 0$.  This gives us two cases, with several sub-cases:
Case $1$:
$$(1-x)(x+3)\le 0\quad\text{and}\quad(x+1)(2-x)> 0$$
Here we need to remember that for a product to be negative the factors have opposite signs, and for a product to be positive then the factors have the same sign.  We have some sub-cases, then:
Case $1a$:
\begin{align*}
1-x &\le 0 \implies x\ge 1\\
x+3 &\ge 0\implies x \ge -3\\
x + 1 &> 0\implies x> -1\\
2 - x &> 0\implies x< 2
\end{align*}
Looking at the overlap between these four inequalities we get $1\le x < 2$.
Case $1b$:
\begin{align*}
1-x &\le 0 \implies x\ge 1\\
x+3 &\ge 0\implies x \ge -3\\
x + 1 &< 0\implies x< -1\\
2 - x &< 0\implies x> 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $1c$:
\begin{align*}
1-x &\ge 0 \implies x\le 1\\
x+3 &\le 0\implies x \le -3\\
x + 1 &> 0\implies x> -1\\
2 - x &> 0\implies x< 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $1d$:
\begin{align*}
1-x &\ge 0 \implies x\le 1\\
x+3 &\le 0\implies x \le -3\\
x + 1 &< 0\implies x< -1\\
2 - x &< 0\implies x> 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $2$:
$$(1-x)(x+3)\ge 0\quad\text{and}\quad(x+1)(2-x)< 0$$
Case $2a$:
\begin{align*}
1-x &\ge 0 \implies x\le 1\\
x+3 &\ge 0\implies x \ge -3\\
x + 1 &> 0\implies x> -1\\
2 - x &< 0\implies x> 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $2b$:
\begin{align*}
1-x &\ge 0 \implies x\le 1\\
x+3 &\ge 0\implies x \ge -3\\
x + 1 &< 0\implies x< -1\\
2 - x &> 0\implies x< 2
\end{align*}
Looking at the overlap between these four inequalities we get $-3\le x < -1$.
Case $2c$:
\begin{align*}
1-x &\le 0 \implies x\ge 1\\
x+3 &\le 0\implies x \le -3\\
x + 1 &> 0\implies x> -1\\
2 - x &< 0\implies x> 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $2d$:
\begin{align*}
1-x &\le 0 \implies x\ge 1\\
x+3 &\le 0\implies x \le -3\\
x + 1 &< 0\implies x< -1\\
2 - x &> 0\implies x< 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Overall, then, we get the solution $-3\le x < -1$ and $1\le x < 2$.
