## Precalculus (6th Edition) Blitzer

$(-\infty, -3) \cup (-3, 5) \cup (5, +\infty)$
Factor the denominator to obtain: $g(x) = \dfrac{3}{(x-5)(x+3)}$ The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression. Look for the real numbers that will make the denominator of the given function equal to zero. These numbers are $5$ and $-3$. Thus, the value of $x$ can be any real number except $5$ and $-3$. Therefore, the domain is $(-\infty, -3) \cup (-3, 5) \cup (5, +\infty)$.