The derivative of a degree $5$ polynomial $p\in\mathbb{C}[z]$ is a degree four polynomial $p'\in\mathbb{C}[z]$, and as such, the zeros of $p'$ may be found explicitly using the quartic formulae.


One may think of a finite Blaschke product as playing the same role on the unit disk that a polynomial plays on the plane (in some ways!). In particular, a degree $5$ finite Blaschke product has $5$ zeros and $4$ critical points in the disk. However in general, a finite Blaschke product $B$ will have additional critical points outside the unit disk.


Suppose $B$ is a finite Blaschke product of degree $n=5$. Let $w\_1,w\_2,w\_3,w\_4\in\mathbb{D}$ be the critical points of $B$ which lie in the disk. Assume that no $w\_i=0$. Due to the conjugate symmetry of $B$, the critical points of $B$ outside the unit disk are exactly $1/\bar{w\_1},1/\bar{w\_2},1/\bar{w\_3},1/\bar{w\_4}\in\mathbb{D}\setminus\overline{\mathbb{C}}$. Thus if $B'=P/Q$ for relatively prime polynomials $P$ and $Q$, then $\deg(P)=8$.


**Can the known conjugate symmetry of the degree $8$ polynomial $P$ across the unit circle allow us to find its zeros using the quartic formulae?**


NOTE 1: The above is a narrowing of [this question](https://math.stackexchange.com/questions/2266593/factoring-real-polynomials-with-no-real-zeros-and-other-polys-whose-zeros-come) on MSE, wherein I also asked about whether a degree $8$ polynomial $p(z)$ whose set of zeros is known to consist of four conjugate pairs could be factored somehow using the quartic formula.


NOTE 2: The same question above goes for degree $4$ finite Blaschke products. Of course for a degree $3$ or less Blaschke product, there are at most $4$ critical points anyway, so one may just use the quartic formula straightforwardly.



Finding the critical points of a degree $5$ Blaschke product
Even for degree $4$ Blaschke products, the critical points need not be contained in a solvable Galois extension of the field of coefficients of $P$. For instance, let $t$ be another variable which is supposed to parametrize the purely imaginary numbers. Take e.g. $w\_k=1/(k+1)+t$. Then the Blaschke product is, up to a constant factor, $B(z)=\prod\frac{z-1/(k+1)-t}{(1/(k+1)-t)z-1}$. The numerator $P(z)$ of the derivative of $B(z)$ is a degree $6$ polynomial in $z$ over the field $\mathbb Q(t)$. It is not hard to see that the Galois group of $P(z)$ over $\mathbb C(t)$ is the symmetric group $S\_6$. So there is no quartic formula expressing the $w\_k$'s in terms of the coefficients of $P(z)$.


In order to get an arithmetic example, note that the Galois group of $P(z)$ over the smaller field $\mathbb Q(\sqrt{-1},t)$ is $S\_6$ even more. So generically (for instance use Hilbert's irreducibility theorem) upon specializing $t$ to $\sqrt{-1}a$ for most rationals $a$ we get that the Galois group of $P(z)$ is still $S\_6$. Instead of using HIT, one may use explicit examples. For instance $a=1/2$ works.


The same kind of construction works for degree $5$ too.


