#### Answer

$\frac{(3n+2)}{(n+4)}$

#### Work Step by Step

Since the numerator consists of a trinomial, we use the rules of factoring trinomials in order to factor it. For the case of the denominator, we use the rule $a^{2}-b^{2}=(a+b)(a-b)$ to factor it. Then, we cancel out the resultant common factors in the numerator and the denominator:
$\frac{3n^{2}-10n-8}{n^{2}-16}$
=$\frac{3n^{2}+2n-12n-8}{n^{2}-4^{2}}$
=$\frac{n(3n+2)-4(3n+2)}{(n+4)(n-4)}$
=$\frac{(3n+2)(n-4)}{(n+4)(n-4)}$
=$\frac{(3n+2)}{(n+4)}$