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

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.

Hint To show that there is at least one nonreal root, we can observe the second derivative of the polynomial $p(x)$ on the l.h.s. is $$p''(x) = 32(3 x^2 - 3 x + 1);$$ this has discriminant $32^2 \cdot (-3) < 0$ and hence has the same sign for all $x$. What does this tell you about the number of real roots $p(x)$ can have?

For the sum, it's easier to work in the variable $x = u - \tfrac{1}{2}$. Then, $$p(x) = q(u) := 8 u^4 + 4 u^2 + \left(k - \tfrac{3}{2}\right) ,$$ which in particular is even. So, if $r$ is a root of $q(u)$, then so is $-r$, and hence $-\frac{1}{2} \pm r$ are roots of $p(x)$.

The I-take-no-clever-shortcuts approach:

I consider the polynomial $p(x)=8x^4-16x^3+16x^2-8x$.

I try to rewrite $p(x)$ as something of the form $d(ax^2+bx+c)^2$ because if it is so, I know how to find the roots. The $d$ coefficient is there only because I like my leading coefficient of $x^4$ to be rational, and $8$ is not a square, but $8/2=4=2^2$ it is. So let me assume $d=2$, which is like to say I'm considering the coefficients of $p(x)/2$

Expanding $(ax^2+bx+c)^2$ I find $$a^2x^4 + 2 a b x^3 +(b^2 + 2 a c) x^2+2 b c x+c^2$$

it is not difficult to see if there is a choice (actually there are two with opposite signs) of $a,b,c$ that makes the above polynomial equal to the source one, minus the constant terms (that we will send to the right anyway).

Indeed we can immediately find that $a=\pm 2$. Let me try with $a=2$. From that I derive (from the coefficient of $x^3$) that $b=-2$ and then $c=1$.

All in all we have obtained $p(x)=2(2x^2-2x+1)^2 -2$ and thus that $p(x)=-k$ can be rewritten as $$ 2(2x^2-2x+1)^2 = 2-k $$

Now, it is not difficult to work out an expression for the roots of this equation. You simply consider the two equations (one for each sign) $$ 2x^2-2x+1 = \pm \sqrt{1-k/2} $$ and solve for $x$ using the classic formula. At the end you obtain four solutions, namely $$ x_{1,2,3,4}=\frac{1}{2}\left(1\pm\sqrt{-1\pm\sqrt{4-2k}}\right) $$

From here, it is easy to answer the questions of the problem.