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

a. $(2t^{2}+5)(4t+7)$ b. In Lesson 8-6, we rewrote the middle term as a sum or difference of two terms, and then we would factor out groups of factors. Here, we already have two middle terms.
a. $8t^{3}+14t^{2}+20t+35=$ $(8t^{3}+14t^{2})+(20t+35)=$ ...factor out the GCF of each group of two terms. $=2t^{2}(4t+7)+5(4t+7)$ ...factor out the common factor, $4t+7$. $=(2t^{2}+5)(4t+7)$ b. In Lesson 8-6, we rewrote the middle term as a sum or difference of two terms, and then we would factor out groups of factors. Here, we already have two middle terms.