jmc

We cannot divide by 0 so we have to exclude from the domain the values of $x$ that make the denominator 0. First we factor the denominator into $(x-3)(x+7)$. Then we set it equal to 0 and solve for $x$. We find that $x$ cannot be 3 or -7, so $x\in \boxed{(-\infty ,-7)\cup (-7,3)\cup (3,\infty )}.$