Odd degree polynomials have a real root

Limits. For $x\ne0$,
$$P(x)=x^n\left(a_n+\frac{a_{n-1}}x+\dots+\frac{a_0}{x^n}\right),$$
and the bracket tends to $a_n$ when $x\to\pm\infty$. Assume for instance $a_n>0$ (otherwise replace $P$ by $-P$). Since $n$ is odd, $x^n\to+\infty$ as $x\to+\infty$ and $x^n\to-\infty$ as $x\to-\infty$. Hence $P(x)\to+\infty$ at $+\infty$ and $P(x)\to-\infty$ at $-\infty$.

Sign change. There exist $A$ and $B$ with $P(A)<0<P(B)$. Since $P$ is continuous, the intermediate value theorem gives a root between $A$ and $B$.

Even degree. The polynomial $x^2+1$ has even degree and no real root: it tends to $+\infty$ at both ends and never changes sign.

Remark. Compare with the previous roots problems: here we do not know any explicit interval, but we know that one exists.