Answer
$\frac{(2x+5)(x-5)}{4}$
Work Step by Step
Factor $x^{2}-8x+15$ as $(x-3)(x-5)$
It can be written as : $\frac{(x-3)(x-5)}{1}$
Factor $\frac{2x+5}{4x-12}$ as $\frac{(2x+5)}{4(x-3)}$
This becomes : $\frac{(2x+5)}{4(x-3)} \times \frac{(x-3)(x-5)}{1}$
After canceling out common factor(s), this becomes:
$\frac{(2x+5)(x-5)}{4}$
