Question1 Is the problem of finding a rational root of $f(x)$ in P for every $n$ (say, for $n=5$)?
Update 1 After I posted the question, I noticed an answer by Robert Israel to the previous question (of Joseph O'Rourke). That could give an answer to my question but I am still not sure how one can avoid factoring numbers $a_0,a_n$.
Update 2 Robert Israel's explanations (more closely related to the original question). Suppose see his comment here ) convince me that all roots his algorithm of $f$ are rational. Can one find a rational root by the following procedure
Step 1. Find checking whether a number $p_1$ that is polynomially dependent on the coefficients of $f$ (say, the discriminant of $f$) and check if $p_1$ polynomial has a rational root $b_1$ of degree $k_1\le n$(all roots rational) runs in polynomial time.
I removed Question 2 . Find a number $p_2$ so that is polynomially dependent on the coefficients of $f$ and $b_1$ and check if it has a rational root of degree $k_2\le n$. I can accept Michael Stoll's answer. ...
The number $p_m$ ($m\le n$) is a rational root of $f$.
Note that existence of such I will post Question 2 as 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 trivialseparate question.