Answer
(7y + 4)($2y^{2} + 1$)
Work Step by Step
$14y^{3} + 8y^{2} + 7y + 4$
Factor by grouping, group the first two terms together.
($14y^{3} + 8y^{2} ) + (7y + 4$)
Factor out the common factors from both parenthesis.
$2y^{2}(7y + 4) + 1(7y + 4$)
We take $(7y + 4$) and factor it out which gives us.
(7y + 4)($2y^{2} + 1$)
