Answer
$(x+4)(x-4)(y-2)$
Work Step by Step
$x^{2}y-16y+32-2x^{2}=$... factor in pairs
$=(x^{2}y-16y)+(32-2x^{2})=$
$=y(x^{2}-16)+2(16-x^{2})$... factor out -1 in the 2nd term
$=y(x^{2}-16)-2(x^{2}-16)$
$=(x^{2}-16)(y-2)$
...The first parentheses hold a difference of squares $(x)^{2}-(4)^{2}$
= $(x+4)(x-4)(y-2)$
