Let $\alpha$ be a root of a polynomial $a\_nx^n + \ldots + a\_1x$ with integral coefficients. 


I would like to determine $\varepsilon > 0$ depending on $a\_1, \ldots, a\_n$ so that $|\alpha| < \varepsilon$ implies $\alpha = 0$.


Is it possible to give a "formula" for such an $\varepsilon$ without refering to the complete list of roots?



Prove that given root of a polynomial is zero by approximation
Of course we must assume some $a\_j \ne 0$. Say $a\_j$ is
the one with least index. Then you want $\varepsilon$ such that 
$p(x) = a\_n x^{n-j} + \ldots + a\_j \ne 0$ for $|x| < \varepsilon$.
You may use inequalities such as
$|p(x)| \ge |a\_j| - \sum\_{k=j+1}^{n} |a\_k| |x|^{k-j} \ge |a\_j| - m \sum\_{k=j+1}^n |a\_k|$


where $m = \max(|x|^{n-j}, |x|)$. Thus $p(x) \ne 0$ if $m < |a\_j|/\sum\_{k=j+1}^n |a\_k|$.


