Show that $8x^4 −16x^3 +16x^2 −8x+k = 0$ has at least one non-real root for all real $k$. Find the sum of the non-real roots 
Show that $8x^4 −16x^3 +16x^2 −8x+k = 0$ has at least one non-real root for all real $k$. Find the sum of the non-real roots. 

Since this polynomial looks so symmetric, I think factoring it might help. We have that $8(x^4-2x^3+2x^2-x) = 8x(x-1)(x^2-x+1)= -k$. Then I'm not sure how to work with the non-real root part, but I think that $(x^2-x+1)$ may have something to do with it.

No need to use derivatives, if you want to continue from your partial attempt.
$(x^2-x)(x^2-x+1) = (x^2-x+\frac{1}{2})^2-\frac{1}{4} = ((x-\frac{1}{2})^2+\frac{1}{4})^2-\frac{1}{4}$.
We can obviously see that it's decreasing for $x < \frac{1}{2}$ and increasing for $x > \frac{1}{2}$.
We certainly know more, since we want to find all $x$ such that:
$((x-\frac{1}{2})^2+\frac{1}{4})^2 = c$
where $c$ is some real.
$(x-\frac{1}{2})^2 = \pm\sqrt{c}-\frac{1}{4}$.
$x = \frac{1}{2} \pm \sqrt{\pm\sqrt{c}-\frac{1}{4}}$. [The signs are independent.]
All roots come in pairs with the outer "$\pm$". So the answer just depends on how many non-real roots. If you want to determine exactly when it has $2$ or $4$ non-real roots, just split into cases based on the $\sqrt{}$, giving the cases $c < 0$, or $0 \le c < \frac{1}{16}$, or $\frac{1}{16} \le c$. You may need to handle the boundaries separately if you don't want to count double roots as separate.
