Answer
$\frac{(3n+5)}{(n-2)}$
Work Step by Step
Since the numerator and the denominator both consist of a trinomial, we use the rules of factoring trinomials in order to factor them. Then, we cancel out the resultant common factors in the numerator and the denominator:
$\frac{9n^{2}+30n+25}{3n^{2}-n-10}$
=$\frac{9n^{2}+15n+15n+25}{3n^{2}+5n-6n-10}$
=$\frac{3n(3n+5)+5(3n+5)}{n(3n+5)-2(3n+5)}$
=$\frac{(3n+5)(3n+5)}{(3n+5)(n-2)}$
=$\frac{(3n+5)}{(n-2)}$
