## Precalculus (6th Edition)

Kurt's answer is correct. The factored form of the given polynomial is $(8a-3)(2a-5)$
Factor the polynomial by grouping: Regroup: $=(16a^2-40a)+(-6a+15)$ Factor out $8a$ in the first and $-3$ in the second group to obtain: $=8a(2a-5)+(-3)(2a-5)$ Factor out $2a-5$ to obtain: $=(2a-5)[8a+(-3)] \\=(2a-5)(8a-3) \\=(8a-3)(2a-5)$ Thus, Kurt's answer is correct.