Answer
$\frac{g(x)}{f(x)} = x + 4$
$\text{The domain is all real numbers except }4$.
Work Step by Step
In this problem, we are asked to divide one function from another. Let us go ahead and set up the division problem:
$\dfrac{g(x)}{f(x)} = \dfrac{x^2 - 16}{x - 4}$
Let's factor the numerator to see if we can cancel one of the factors with the denominator. We can see that the binomial in the numerator is the difference of two squares. The factorization for the difference of two squares is as follows:
$(a^2 - b^2) = (a - b)(a + b)$
Let us plug in $\sqrt {x^2}$ or $x$ for $a$ and $\sqrt {16}$ or $4$ for $b$:
$\dfrac{g(x)}{f(x)} = \dfrac{(x - 4)(x + 4)}{x - 4}$
We can cancel out an $x - 4$ from both the numerator and denominator:
$\dfrac{g(x)}{f(x)} = x + 4$
We cannot have $x = 4$ because if $x$ were $4$, the denominator would be $0$. The denominator cannot equal $0$ or else the function would be undefined.
The domain is all real numbers except $4$.