Answer
$\frac{(3a+2)}{(4a-1)}$
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:
$\frac{6a^{2}-11a-10}{8a^{2}-22a+5}$
=$\frac{6a^{2}+4a-15a-10}{8a^{2}-2a-20a+5}$
=$\frac{2a(3a+2)-5(3a+2)}{2a(4a-1)-5(4a-1)}$
=$\frac{(3a+2)(2a-5)}{(4a-1)(2a-5)}$
=$\frac{(3a+2)}{(4a-1)}$
