Answer
$\color{blue}{(-\infty, -3) \cup (-3, -2) \cup (-2, +\infty)}$
Work Step by Step
Factor the denominator to obtain:
$=\dfrac{12}{(x+3)(x+2)}$
The denominator of a rational expression is not allowed to be equal to zero as it will make the expression undefined.
Find the values of $x$ for which the denominator is equal to zero by equating the denominator to zero:
$(x+3)(x+2)= 0$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
\begin{array}{ccc}
&x+3 = 0 &\text{ or } &x+2=0
\\&x=-3 &\text{or} &x=-2
\end{array}
This means that the value of $x$ can be any real number except $-3$ and $-2$.
Therefore, the domain of the given rational expression is the set of real numbers except $-3$ and $-2$.
In interval notation, the domain is:
$\color{blue}{(-\infty, -3) \cup (-3, -2) \cup (-2, +\infty)}$
