Prove that a polynomial of odd degree in $ \Bbb R[x]$ with no multiples roots must have an odd number of real roots.

Question

The coefficients of the polynomial are in the ring of real numbers. Prove that a polynomial of odd degree in $ \Bbb R[x]\\$ with no multiple roots must have an off number of real roots. I hate to ask people to solve things, so I was wondering if you could glance at my attempt and give a suggestion. There is a theorem in my textbook that states that every polynomial of odd degree has a root.

Let f(x) be a polynomial of odd degree. Since, I know that since f(x) is odd it has a root, and thus can be factored as follows: Let a be an arbitrary root of f(x). With a $\in \Bbb R$, then f (x) = (x-a)(h(x)) where h(x) $\in \Bbb R[x]$. This implies that h(x) must have even degree, because deg(f(x)) is odd and deg (x-a) is odd. At this point can I assume that, since h(x) must have even degree, it must have an even number of roots and therefore factors? Because, to the contrary, if it had an odd number of factors or roots, our initial assumption that f(x) has odd degree would violated?

Answer 1

Hint: Any roots that aren't real will occur in pairs (why?)

Answer 2

The only deep result you need is

An odd-degree polynomial $f(x)\in\mathbb{R}[x]$ has at least a root.

Also you can check algebraically whether $f$ has multiple roots (in some extension field), by computing the greatest common divisor between $f$ and its derivative.

Another easy observation is that if $u\in\mathbb{C}$ is a root of $f$, then also $\bar{u}$ is a root, which is distinct from $u$ if $u\notin\mathbb{R}$.

Thus you can factor your polynomial as $$ (x-a_1)\dots(x-a_r) (x-u_1)(x-\bar{u}_1)\dots(x-u_s)(x-\bar{u}_s) g(x) $$ where $a_1,\dots a_r$ are the real roots and $u_1,\bar{u}_1,\dots,u_s,\bar{u}_s$ are the complex (nonreal) roots, and $g$ has no root (either real or complex). Note that $g$ must have even degree, otherwise it would have a real root.

Now count degrees.

Note. The FTA says that $g$ has degree zero, but this is not needed.