TITLE: Proof that every polynomial of odd degree has one real root
QUESTION [32 upvotes]: I want to prove that every real polynomial of odd degree has at least one real root, using  the intermediate value theorem.
Let $P(x) = x^{2n+1} + a_n x^{2n} + . . . + a_0$ for each $a_i \in \mathbb{R}$ and $n \in \mathbb{N}$.
By the fundamental theorem of algebra I know that $P(x)$ has exactly $2n+1$ complex roots, so
$P(x) = (x+r_1)(x+r_2) . . . (x+r_{2n+1})$ for each $r_i \in \mathbb{C}$
I do not know how to complete this but I do know that, at some point, I probably have to show that each root with imaginary part non zero has to come in conjugate pairs, and since $2n+1$ is odd there is at least $1$ root that is imaginary part $0$ and thus real.

REPLY [11 votes]: Let $p(x)=a_0 + a_1 x + \dots + a_n x^n$ a polynomial $p: \mathbb{R} \to \mathbb{R}$ with $n$ odd and $a_n \neq 0$. Suppose $a_n >0$. We can write $p(x) = a_n x^n \cdot r(x)$ with
$$r(x) = \frac{a_0}{a_n} \cdot \frac{1}{x^n} + \frac{a_1}{a_n} \cdot \frac{1}{x^{n-1}}+ \cdots + \frac{a_{n-1}}{a_n} \cdot \frac{1}{x} + 1.$$
So, we have
$$\lim_{x \to +\infty} r(x) = \lim_{x \to -\infty} = 1.$$
Thus
$$\lim_{x \to +\infty} p(x) = \lim_{x \to +\infty} a_n x^n = +\infty$$
and
$$\lim_{x \to -\infty} p(x) = \lim_{x \to -\infty} a_n x^n = -\infty$$
because $n$ is odd. Therefore the interval $p(\mathbb{R})$ is ilimited inferior and superiorly, i.e. $p(\mathbb{R}) = \mathbb{R}$. This means that $p: \mathbb{R} \to \mathbb{R}$ is sujective. In particular, there is $c \in \mathbb{R}$ that $p(c)=0$. 
For a polynomial with even degree, take $p(x)=x^2+1$. This have not real roots.