Answer
$(3y+4)(2y-1)(2y+1)$
Work Step by Step
$ 12y^{3}+16y^{2}-3y-4\qquad$...factor in groups.
$=(12y^{3}+16y^{2})+(-3y-4)$
$=4y^{2}(3y+4)-(3y+4)\qquad$...factor out the common binomial factor,$3y+4$.
$=(3y+4)(4y^{2}-1)\qquad$...recognize the difference of two squares: $a^{2}-b^{2}=(a-b)(a+b)$
$=(3y+4)(2y-1)(2y+1)$
