This question is related to this question of Joseph O'Rourke and this question of mine.

** Question.** Let $f$ be a polynomial with integer coefficients. Suppose that all roots of $f$ are rational. Can one find a rational root by the following procedure (similar to solving equations in radicals):

Step 1. Find a number $p_1$ that is polynomially dependent on the coefficients of $f$ (say, the discriminant of $f$) and check if $p_1$ has a rational root $b_1$ of degree $k_1\le n$. Thus $p_1=g_1(a_0,...,a_n)$ for some polynomial $g_1$, and $b_1^{k_1}=p_1$.

Step 2. Find a number $p_2$ that is polynomially dependent on the coefficients of $f$ and $b_1$ and check if it has a rational root of degree $k_2\le n$. Thus $p_2=g_2(a_0,...,a_n,b_1)$ for some polynomial $g_2$, and $b_2^{k_2}=p_2$.

....

The number $p_m$ ($m$ polynomially depends on $n$) is a rational root of $f$. Of course all polynomials $g_i$ involved in this procedure should not depend on the coefficients of the polynomial $f$, only on its degree (like the discriminant).

Note that existence of such a procedure does not contradict unsolvability of the symmetric group $S_n$, $n\ge 5$, because we assume that the Galois group of $f$ is trivial.