## Algebra: A Combined Approach (4th Edition)

$(y+2)(y-2)(y^2+4)$
$y^4 - 16 = (y^2)^2 - (4)^2 = (y^2 - 4)(y^2 + 4)$ and then $y^2 - 4 = (y)^2 - (2)^2 = (y-2)(y+2)$ Therefore $y^4 - 16 = (y^2)^2 - (4)^2 = (y^2 - 4)(y^2 + 4) = (y+2)(y-2)(y^2+4)$