Chapter 8 - Polynomials and Factoring - 8-8 Factoring by Grouping - Practice and Problem-Solving Exercises - Page 520: 22

$(8p^{2}+28)(p-4)$

$8p^{3}=2.2.2.p.p.p$ and $32p^{2}=2.2.2.2.2.p.p$ Hence, $GCF=$$2.2.2.p.p=8p^{2}$ $28p = 2.2.7.p$ and $112=2.2.2.2.7$ Hence, $GCF=$$2.2.7=28$ After grouping: $=(8p^{3}-32p^{2})+(28p-112)$ $=8p^{2}(p-4)+28(p-4)$ $=(8p^{2}+28)(p-4)$