polynomial equation of odd degree

Theorem.

The equation

$a_0{x}^n+a_1{x}^{n-1}+\mathrm{\cdots }+a_{n-1}x+a_n=0$(1)

with odd degree $n$ and real coefficients $a_i$ ($a_0\ne 0$) has at least one real root $x$.

Proof.  Denote by $f(x)$ the left hand side of (1).  We can write

$$f(x)=a_0{x}^n[1+g(x)]$$

where  $g(x):={\displaystyle \frac{a_1}{x}}+\mathrm{\cdots }+{\displaystyle \frac{a_{n-1}}{x^{n-1}}}+{\displaystyle \frac{a_n}{x^n}}$.  But we have $\underset{|x|\to \mathrm{\infty }}{lim}g(x)=0$  because

$$\underset{|x|\to \mathrm{\infty }}{lim}\frac{a_i}{x^i}=0$$

for all  $i=1,\mathrm{\dots },n$.  Thus there exists an  $M>0$  such that

Accordingly  $1+g(\pm M)>0$  and

$$\text{sign}f(\pm M)=(\text{sign}{a}_0){(\text{sign}(\pm M))}^n\cdot 1=(\text{sign}{a}_0)(\pm 1)$$

since $n$ is odd.  Therefore the real polynomial function $f$ has opposite signs in the end points of the interval  $[-M,M]$.  Thus the continuity of $f$ guarantees, according to Bolzano’s theorem, at least one zero $x$ of $f$ in that interval.  So (1) has at least one real root $x$.