Prove that odd polynomials have at least one root. Intuitively this is easy for me to understand, but I don't know how to start the proof. Can someone help me? 

Prove that any polynomial of the form $x^n+\sum_{i<n}a_{i}x_{i}$, with $a_{i}\in \mathbb{R}$ and $n$ odd, has at least one real root.

 A: Use Intermediate Value Theorem.
Since $\lim_{x\to\infty}f(x)=\infty$, and $\lim_{x\to-\infty}f(x)=-\infty$, in particular there exists $(a,b)$ such that $f(b)>0$ and $f(a)<0$.
By IVT, there exists $c\in (a,b)$ such that $f(c)=0$.
A: If
$f(x)
= x^n+\sum_{i<n}a_{i}x^{i}
$,
(you wrote $x_i$;
I assume you meant
$x^i$)
then
for large enough $x$,
$f(x) > 0$.
To show this,
write
$f(x)
= x^n(1+\sum_{i<n}a_{i}x^{i-n})
$.
We want to choose $x$ large enough
so that
$\sum_{i<n}|a_{i}|x^{i-n}
< 1$.
This will be true if
$|a_{i}|x^{i-n}
< \frac1n
$
or
$x^{n-i} > n|a_{i}|
$
or
$x > (n|a_{i}|)^{1/(n-i)}
$.
Therefore,
if
$x > \max_{i<n} (n|a_{i}|)^{1/(n-i)}
$,
$f(x) > 0$.
Note:
There are other, better bounds
that can be worked out.
This is an easy one.
This is true whether
$n$ is even or odd.
If $n$ is odd,
then
$x^n < 0$ for $x < 0$.
Therefore,
if $x < - \max_{i<n} (n|a_{i}|)^{1/(n-i)}$,
then,
as before,
$\sum_{i<n}|a_{i}||x|^{i-n}
< 1$
so that
$|\sum_{i<n}a_{i}x^{i-n}|
\le \sum_{i<n}|a_{i}x^{i-n}|
< 1$
so that
$1+\sum_{i<n}a_{i}x^{i-n}
> 0$.
Therefore,
if $x < - \max_{i<n} (n|a_{i}|)^{1/(n-i)}$,
then
$f(x)
= x^n(1+\sum_{i<n}a_{i}x^{i-n})
< 0
$.
Since $f(x) > 0$
for all large enough $x$
and
$f(x) < 0$
for all negatively large $x$
(with
$|x| > \max_{i<n} (n|a_{i}|)^{1/(n-i)}$
working in both directions),
since $f(x)$ is continuous,
there has to be an $x_0$
such that
$f(x_0) = 0$.
