## Algebra 1: Common Core (15th Edition)

$\frac{(2x+5)(x-5)}{4}$
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: $\frac{(2x+5)(x-5)}{4}$