algebra precalculus - Polynomial with odd number of real roots

I have been trying to characterize the number of roots on $\theta$ for the following polynomial

$$\sum_{i=1}^n \frac{\theta- x_i}{1+ \left(x_i - \theta \right)^2} = 0$$

If we were to put everything under a common denominator then we would see that the polynomial is of degree $2n -1$ with coefficients depending on the $x_i$ for $i=1,2,...,n$ . Taking limits as $\theta \to \infty$ and $\theta \to -\infty$ we find that the polynomial has at least one real root as it tends to zero through positive and negative values respectively.

What I do not understand is how one reaches the conclusion that the number of real roots is odd in this case. I know that the complex roots are $2 n -1$ but I'm talking about the real roots. Is there a theorem that guarantees that or is that common sense hiding in plain sight, at least for me?

Thank you.

Answer

If you write your equation in the form of $p(\theta)=\theta^m+a_{m-1}\theta^{m-1}+...+a_1\theta+a_0=0$, you get that $m=2n+1$ is an odd number and $a_i\in \mathbb R$. So, $p(\theta)$ is a polynomial with real coefficient by odd degree. Thus, $p(\theta)$ has at least one real root. But if $z\in \mathbb C$ is a root of a polynomial with real, then $\overline z$ is to. So the number of complex root of any polynomial with real coefficient, is even. From here, the number of real root of a polynomial with real coefficient, is odd.