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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?

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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.

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The conditions of the first question is the typical for the polynomial equations with real coeficients, so the first answer is negative.

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 $$z = z_k \exp{2\pi n\over deg},\quad n\in\{0\dots deg-1\}.$$

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