4.
\(\frac{f}{g}\,=\,\frac{f(x)}{g(x)} \\~\\ \frac{f}{g}\,=\,\frac{x^2-16}{x+4}\)
Before simplifying, let's get the domain. x + 4 is in a denominator, so...
x + 4 β 0
x β -4 The domain is all real numbers with the exception that x β -4
\(\frac{f}{g}\,=\,\frac{x^2-16}{x+4}\)
Factor the numerator as a difference of squares.
\(\frac{f}{g}\,=\,\frac{(x-4)(x+4)}{x+4}\)
Reduce the fraction by (x + 4) .
\(\frac{f}{g}\,=\,\frac{(x-4)}{1} \\~\\ \frac{f}{g}\,=\,x-4\)
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