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

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?