Intermediate Value Theorem and polynomial of odd degree As a consequence of Intermediate Value Theorem (IVT), it is illustrated in almost all calculus books and notes/courses that every polynomial of odd degree over real numbers have a real root.
I tried to see whether this result has been proved without the use of intermediate value property, and I did not find any link or question related to that on this and other sites.
Can one suggest me whether there is a proof of above mentioned consequence of IVT, which is done without using IVT?

I had seen that the fundamental theorem of algebra have been of interest to many people for proving just using algebraic techniques, but among all the known various proofs of FTA, epsilon amount of analysis is used.
So, my above question is of similar nature, but I did not see any comments on its proofs without IVT.
 A: Complex roots of a polynomial with real coefficients occur in pairs.
Hence?
https://en.m.wikipedia.org/wiki/Complex_conjugate_root_theorem
A: Every polynomial $P$ can be split (not sure of the word, the one in French is "scindé") in $\mathbb{R}$ : $P = \prod (X^2-a_k X + b_k) \prod (X- \lambda_k)$. As the degree of $P$ is odd, there is at least one factor $(X-\lambda)$, so there is at least one real root.
As for Vivid's proof, the proof of Descarte's Rule of Signs is based on IVT, so it is not really a different proof.

(Sorry if my words are incorrect, I'm French and I'm not used to writing maths in English.)

A: One could use Descarte's Rule of Signs. So, below is the idea:
Assume that we have $$P(x) = \sum_{k=0}^{n} a_kx^k$$

*

*If there is at least one sign change, then we are done. So we assume there is not.

*Assume there are terms of both odd and even exponents, then we analyze $P(-x)$. There were no sign changes in $P(x)$, therefore, we must have in $P(-x)$ since $(-x)^{\text{odd}} = -x^{\text{odd}}$ and $(-x)^{\text{even}} = x^{\text{even}}$.

*Assume there are only odd powers, then the polynomial is an odd function which means $x=0$ is a root.

*We cannot assume all exponents are even since $P(x)$ is of odd degree.

