Answer
$(-4, 3]$
Work Step by Step
$\frac{x^3 + 3x^2 - 9x - 27}{x+4} \leq 0$
$\frac{(x^2 - 9)(x + 3)}{x+4} \leq 0$
$\frac{(x - 3)(x + 3)^2}{x+4} \leq 0$
Find the zeros of the expressions in the numerator AND the denominator
$x = 3, -3, -4$
Test numbers in between those zero values to determine if the function is negative or positive
$(-∞, -4)$ $\frac{(-)(+))}{(-)} = (+)$
$(-4, -3]$ $\frac{(-)(+)}{(+)} = (-)$
$([-3, 3]$ $\frac{(-)(+)}{(+)} = (-)$
$[3, ∞)$ $\frac{(+)}{(+)(+)} = (+)$
Thus the solution is $(-4, 3]$
