Algebra 1

$(3y+1)(4y^{2}-3)$
$12y^{3}=2.2.3.y.y.y$ and $4y^{2}=2.2.y.y$ Hence, $GCF=$$2.2.y.y=4y^{2} 9y=$$3.3.y$ and $3=3.1$ Hence, $GCF=3$ After grouping; $=(12y^{3}+4y^{2})-(9y+3)$ $=4y^{2}(3y+1)-3(3y+1)$ $=(3y+1)(4y^{2}-3)$