How do you factor 8x^6-32x^5+4x^4?

$y = 4 {x}^{4.} \left(2 {x}^{2} - 8 x + 1\right)$
The trinomial in parentheses can't be factored because $D = {b}^{2} - 4 a c = 64 - 8 = 56$ is not a perfect square.