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