Answer
$4y(y+2)(3y+1)$
Work Step by Step
$ 12y^{3}+28y^{2}+8y\qquad$...factor out the common term, $4y$.
$=4y(3y^{2}+7y+2)$
... Searching for two factors of $ac=6$ whose sum is $b=7,$
we find$\qquad 6$ and $1.$
Rewrite the middle term and factor in pairs:
$=4y(3y^{2}+6y+y+2)=$
$=4y[3y(y+2)+(y+2)]$
=$4y(y+2)(3y+1)$
