## Elementary Algebra

$\frac{(2n-3)}{(n+1)}$
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: $\frac{4n^{2}-12n+9}{2n^{2}-n-3}$ =$\frac{4n^{2}-6n-6n+9}{2n^{2}+2n-3n-3}$ =$\frac{2n(2n-3)-3(2n-3)}{2n(n+1)-3(n+1)}$ =$\frac{(2n-3)(2n-3)}{(n+1)(2n-3)}$ =$\frac{(2n-3)}{(n+1)}$