Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that the Descartes's Criterion is correct. Descartes's criterion: If $a_nx^n + a_{n-1}x^{n-1}+...+a_0$ has a rational root $x = s/t$, where $s$ and $t$ are relatively prime, then t divides $a_n$ and $s$ divides $a_0$.

The hint says that I should factor $a_n$ and $a_0$ to get all possible values of $s/t$ and substitute to find which,if any, is a root. But I think that this hint only works for specific polynomials, not to prove the general statement.

share|cite|improve this question

1 Answer 1

up vote 2 down vote accepted

It does seem that your hint is more about how to apply Descartes' criterion than how to prove it's valid. Here's an approach to the latter: suppose $\frac{s}{t}$ is a root of our polynomial $A(x)=a_nx^n+...+a_0,$ and let's say we've expressed the root in lowest terms, so $s$ and $t$ have no common factors.

So $$a_n\frac{s^n}{t^n}+a_{n-1}\frac{s^{n-1}}{t^{n-1}}+...+a_0=0$$ Now multiply through by $t^n$, and we get $a_ns^n+a_{n-1}ts^{n-1}...+a_1st^{n-1}+a_0t^n=0$. From here you should be able to argue that $s$ and $t$ must both divide the left-hand side and thus $a_n$ and $a_0$ respectively, as desired. Do you see how to do so?

share|cite|improve this answer
Yes, I see it. Thank you. –  user40105 Sep 30 '12 at 2:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.