I have been curious about that for some time now, without finding a proof:
It is well-known that a odd degree polynomial $P(x)$ with real coefficients have at least one root. The usual proof use the fact that $\lim_{x\to\infty}P(x)=\pm\infty$ and $\lim_{x\to-\infty}P(x)=\mp\infty$ and the Intermediate Value Theorem.
On the other hand, Descartes' Rule of Signs (DROS) goes like this:
Let $P(x)$ be a polynomial of degree $n$ with real coefficients. Then:
- The number of positive roots of $P$ is equal to the number $v$ of variations in sign of $P(x)$, or is less than $v$ by an even number.
- The number of negative roots of $P$ is equal to the number $w$ of variations in sign of $P(-x)$, or is less than $w$ by an even number.
Note: A variation in sign occurs when two consecutive coefficients have an opposite sign. We ignore zero coefficients and roots are counted with multiplicity.
I was wondering if I can use the DROS to prove that odd degree polynomials with real coefficients always have a root. The structure of the proof would be something like that:
- If $P(0)=0$, we are done. Assume $P(0)\neq 0$.
- Assume that $P$ has no positive root. Then the number $v$ of variations in sign of $P(x)$ is even.
- Discussion to show that the number $w$ of variations in sign has to be odd, so that $P$ need to have a (negative) root.
Unfortunately, I have been unable to complete the discussion and the proof.
Any idea? Any proof, hint or reference is welcome.