On the lecture note it is written that if a complex polynomial had the roots $z$ and $\overline{z}$ so it can be written as a product of a real polynomial and a complex constant.
I may have copied or misunderstood, but isn't it obvious? what is the purpose of this statement?
On
If polynomial $f(x)$ has only the roots $z$ and $\bar z$ (and with simple multiplicity, and we assume $z\ne \bar z$), then clearly $$ f(x)=c\cdot (x-z)\cdot (x-\bar z) = c\cdot (x^2-(z+\bar z) x+z\bar z)$$ with $c\ne 0$. As $z+\bar z=2\Re z\in\Bbb R$ and $z\bar z=\|z\|^2\in\Bbb R$, the claim follows.
It is not deep, but still worth mentioning.
Probably what was meant was that if the non-real roots of the non-zero polynomial $f(z)$ come in pairs $z$, $\overline{z}$ with each of $z$ and $\overline{z}$ having the same multiplicity, then the polynomial has the form:
$$ f(z) = c(z - x_1)^{l_1}\ldots (z - x_m)^{l_m}[(z - z_1)(z - \overline{z_1})]^{m_1}\ldots[(z - z_k)(z - \overline{z_k})]^{m_k} $$
for some non-zero $c \in \Bbb{C}$, where the $x_i$ are the real roots, if any. If you expand the quadratics in the square brackets, you will find that the coefficients are the real numbers $z_i + \overline{z_i}$ and $z_i\overline{z_i}$, so the coefficients of $f(z)/c$ are real.
This description characterises polynomials with real coefficients. If you think it's obvious, then good for you. The purpose of stating it in your notes is presumably to help people whose algebraic intuitions aren't as good as yours.