# Existence of a real root for polynomials of odd degree

The fundamental theorem of algebra famously states that a polynomial of degree $n$ with real coefficients, say $p \in \mathbb{R}[x]$, has $n$ complex roots when counting multiplicities. This is equivalent to saying $p(x) = \prod_{k=1}^{k=m} (x - \alpha_k)^{h_k}$ where $h_1 + ... + h_m = n$ and $\alpha_i \neq \alpha_j \in \mathbb{C}$ for all $i,j = 1,...,m$ such that $i \neq j$. This is due to the ‘equivalence’ of roots and linear factors.

We will probably get round to proving this at some point so if you want to find out how don’t forget to follow us! Or just google how to prove the fundamental theorem of algebra – It is really up to you.

What is rather nice is that by throwing in the condition that $n$ is odd we can guarantee the existence of a real root. The idea is nicely summed up in the Numberphile video:

Presumably if you watched the video you probably have a very good idea of how to prove the result but let us write it out in mathematics.

We have $p \in \mathbb{R}[x]$ with degree $n$ being odd and suppose $p(x) = a_0 +a_1x + ... a_nx^n$ where $a_i \in \mathbb{R}$ and $a_n \neq 0$. Polynomials are continuous so we must have $p$ being continuous. Looking at the behaviour of $\frac{p(x)}{x^n} = \frac{a_o + ... + a_n x^n}{x^n} = \frac{a_n}{x^n} + ... + a_n$ as $|x| \longrightarrow \infty$ we have $\frac{p(x)}{x^n} \longrightarrow a_n$. If we change $x$ to $-x$, due to $n$ being odd, $x^n$ changes sign. The sign of $\frac{p(x)}{x^n}$ for sufficiently large $x$ always has the same sign as $a_n$ so putting the two facts together means $p(x)$ and $p(-x)$ differ in sign – This is the ‘crossing the x-axis’. We know $p$ is continuous so it is nice and in particular we can use the Intermediate Value Theorem. As $0$ is in the image of $p$ when we take $|x|$ sufficiently large there must be an $\alpha \in \mathbb{R}$ such that $p(\alpha) = 0$ by the intermediate value theorem.