Answer
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.
Work Step by Step
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.
