# Factoring real polynomials with no real zeros, and other polys whose zeros come in pairs

Of course a polynomial of degree at most $4$ may "easily" be factored. And for a polynomial of degree $5$ or greater, no algebraic formula for the roots need exist. What about when the zeros of the polynomial are known to come in conjugate pairs: Suppose that $p(z)$ is a real polynomial (ie $p(t)\in\mathbb{R}$ for all $t\in\mathbb{R}$), but $p$ is non-zero on $\mathbb{R}$ (so that the zeros of $p$ come in conjugate pairs). If $\deg(p)=8$ (or $6$), can the quartic (or cubic) formulas be used to find the zeros of $p$?

Second and related question: suppose that $q$ has no zeros on the unit circle $\mathbb{T}$, and it is known that the zeros of $q$ are conjugate symmetric across the unit circle. (Note, this is the case for the numerator of the derivative of a finite Blaschke product, and this is in fact the motivation for the first question as well.) If $\deg(q)\leq8$, can $q$ be factored somehow using the quartic formula?

The answer to your first question is NO. Consider for example $p(x)=(x-1)^6+x^2+1$. Then PARI/GP tells us that the Galois group of this polynomial is $S_6$, which is not a solvable group, so $P$ cannot be solved by radicals.
The conditions of the second answer corresponds with the equation $$z^{deg} = z_k^{deg},$$ where $z^k$ is anyone of the given roots. The solutions are $$z = z_k \exp{2\pi n\over deg},\quad n\in\{0\dots deg-1\}.$$