This question is related to  [this question](https://mathoverflow.net/questions/97307/polynomials-all-of-whose-roots-are-rational) of Joseph O'Rourke and  [this question](https://mathoverflow.net/questions/97329/complexity-of-finding-a-rational-root-of-a-polynomial)  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. 



Solving polynomial equations in radicals provided all roots are rational 
No.


Consider a generic polynomial $x^n+a\_1x^{n-1}+a\_2x^{n-2}+....+a\_n$ over $\mathbb Q(a\_1,...,a\_n)$. Adjoin $b\_1$, a root of $p\_1$ of degree $k\_1$, then adjoin $b\_2$, and so on. This is contained in some solvable galois extension of $\mathbb Q(a\_1,...,a\_n)$.


Let $q$ be the generic polynomial evaluated at $p\_m$, and let $r$ be the norm to $\mathbb Q(a\_1,...,a\_n)$ of $q$. Then $r$ lies is $\mathbb Q(a\_1,...,a\_n)$, and is zero whenever the polynomial has all rational roots. But this is a Zariski dense subset, so $r$ is identically zero, so this solvable Galois extension always contains a root, which does violate unsolvability of the symmetric group.


