<s>[INST] How do you find the imaginary roots of a fourth degree polynomial that cannot be simplified? I started out with $f(x)=16x^6-1$, and I got it down to $64x^4+16x^2+4$ by synthetically dividing by roots $0.5$ and $-0.5$ How should I continue in order to find the other roots?
 [/INST] Note that $f\left(\pm \frac{1}{2}\right) = 16\left(\pm \frac{1}{2}\right)^6 - 1 = -\frac{3}{4}$, so $\pm \frac{1}{2}$ are not roots of $f$.
One way to proceed with the original problem is to write $v^6 := 16 x^6$, that is, $v = \sqrt[3]{4} x$, so that the original polynomial is $v^6 - 1$. This clearly has roots $v = \pm 1$, which (substituting) gives corresponds to
$$x = \pm 4^{-1/3}.$$ Using synthetic division gives
$$v^6 - 1 = (v - 1)(v + 1)(v^4 + v^2 + 1).$$
One can find the roots of the latter by writing $w = v^2$ and applying the quadratic formula to the resulting expression $w^2 + w + 1$; this only has complex roots, $w = -\frac{1}{2} \pm \frac{\sqrt{3}}{2} i,$ and we can then solve readily for $v$ and then $x$. 
We can also attempt to factor $v^4 + v^2 + 1$. By the Rational Root Theorem it has no roots in $\mathbb{Q}$, so if it has a nontrivial factorization, it must be into a product of quadratics. One can set this up manually---writing down a product of general quadratics, distributing, and equating coefficients---and find the factorization
$$(v^2 + v + 1)(v^2 - v + 1).$$
(On the other hand, if one knows about the cyclotomic polynomials $\Phi_n$, one can write down the factorization of any polynomial $v^m - 1$ as a product of such; in our case,
$$v^6 - 1 = \Phi_1(v) \Phi_2(v) \Phi_3(v) \Phi_6(v) = (v - 1) (v + 1) (v^2 + v + 1) (v^2 - v + 1).)$$
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