## Elementary Algebra

$\frac{(2n-1)}{(n-3)}$
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: $\frac{2n^{2}+5n-3}{n^{2}-9}$ =$\frac{2n^{2}-1n+6n-3}{n^{2}-3^{2}}$ =$\frac{n(2n-1)+3(2n-1)}{(n+3)(n-3)}$ =$\frac{(2n-1)(n+3)}{(n+3)(n-3)}$ =$\frac{(2n-1)}{(n-3)}$