## Algebra 1

(7y + 4)($2y^{2} + 1$)
$14y^{3} + 8y^{2} + 7y + 4$ 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)($2y^{2} + 1$)