Proving a statement involving the rational root theorem The rational root theorem states that, given $P \in \mathbb{Z}[x], P = \displaystyle\sum_{j=0}^n a_j x^j$ with $a_0 \neq 0$ and $a_n \neq 0$, if $P(p/q)=0, p\perp q$ then $p|a_0$ and $q|a_n$.

As a follow-up to this question, I'd like to know how can I prove the following:

$$a_0 = a_n \displaystyle\prod_{j=1}^n (-r_j)$$, with $r_j$ being roots of P(x)

This is the same as saying:

$$a_0 = a_n \frac{-p}{q} \displaystyle\prod_{j=2}^n (-r_j)$$

Rearranging this, we get:

$$a_n = q\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)}$$

I want to prove that $$\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)} \in \mathbb{Z}$$ without using the rational root theorem (actually, I want to prove the rational root theorem using this identity).

So far I've tried to prove it by contradiction, assuming $a_0 = k(-p)\displaystyle\prod_{j=2}^n (-r_j) + \epsilon$ with $\epsilon \in (0, (-p)\displaystyle\prod_{j=2}^n (-r_j))$ but I don't seem to find anything weird rearranging that equation.

If you manage to find a proof please just post a small hint. Thanks in advance.
Those are just Vieta's formulas for the coefficients in terms of the roots.
Those are just Vieta's formulas for the coefficients in terms of the roots.