Answer
$(-4,3]$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle \frac{(x^3+3x^2-9x-27)}{(x+4)} \leq 0$
$f(x)=\displaystyle \frac{(x-3)(x+3)(x+3)}{(x+4)} \leq 0$
2. The cut points are:
$\displaystyle \displaystyle \frac{(x-3)(x+3)(x+3)}{(x+4)} = 0$
$x=-4$ or $x=-3$ or $x=3$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\
& & \displaystyle \displaystyle \frac{(a-3)(a+3)(a+3)}{(a+4)} & \\
(-\infty,-4) & -5 & \frac{(-)(-)(-)}{(-)}=(+) & F\\
(-4, -3) & -\frac{7}{2} & \frac{(-)(-)(-)}{(+)}=(-) & T\\
(-3, 3) & 0 & \frac{(-)(+)(+)}{(+)}=(-) & T\\
(3,\infty) & 10 & \frac{(+)(+)(+)}{(+)}=(+) & F\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-4,3]$
