polynomial equation of odd degree

Theorem 1   The equation

(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 $$f(x) = a_0x^n[1+g(x)]$$ where $\displaystyle g(x) := \frac{a_1}{x}\!+\cdots\!+\!\frac{a_{n-1}}{x^{n-1}}\!+\!\frac{a_n}{x^n}$ . But we have $\displaystyle\lim_{|x|\to\infty}g(x) = 0$ because $$\lim_{|x|\to\infty}\frac{a_i}{x^i} = 0$$ for all $i = 1,\,...,\,n$ . Thus there exists an $M > 0$ such that $$|g(x)| < 1\,\, \mbox{for}\,\, |x| \geqq M.$$ Accordingly $1+g(\pm M) > 0$ and $$\mbox{sign} f(\pm M) = (\mbox{sign} a_0)(\mbox{sign}(\pm M))^n\cdot 1 = (\mbox{sign} a_0)(\pm 1)$$ 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 $x$ .