Question 977700
{{{x^4-256=0}}}
{{{(x^2-16)(x^2+16)=0}}}
{{{(x-4)(x+4)(x^2+16)=0}}}


continuing with the irreducible quadratic,
{{{x^2=-16}}}
{{{x=0+- 4i}}}


The roots for q(x) are  4, -4, and -4i and 4i.