Answer
$2(2y+1)(8y-3)$
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
The square of a binomial sum is: $(A+B)^2=A^2+2AB+B^2$.
The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$.
Hence here: $32y^2+4y-6=\\=32y^2+16y-12y-6\\=16y(2y+1)-6(2y+1)\\=(2y+1)(16y-6)\\=(2y+1)2(8y-3)\\=2(2y+1)(8y-3)$
