Apologies for the simplicity of this question, I just want to make sure that I have this absolutely right.
There are lots of questions on Vieta's formulas on Mathematics Stack Exchange already. But I got confused by this page (here, not a Stack Exchange page), which seems to imply that the formulas offer a process of approximation rather than an accurate series of equalities. I've probably misunderstood the context. However:
Define a polynomial $P(x)=\sum_{i=0}^n a_ix^i$ with rational coefficients and roots (not necessarily distinct) $r_1,r_2,...,r_n$. Vieta's formulas give
$$\sum_{i=0}^nr_i=-\frac{a_{n-1}}{a_n}$$ $$\prod_{i=0}^nr_i=(-1)^n\frac{a_{0}}{a_n}$$
In the case of a monic polynomial $M(x)$ with rational coefficients, we have $a_n=1$ and therefore
$$\sum_{i=0}^nr_i=-a_{n-1}$$ $$\prod_{i=0}^nr_i=(-1)^na_{0}$$
My questions are: (a) do I have this right, and (b) are these precise and 100% correct formulas rather than approximations?
ADDED DETAILS:
Say, for example, that I want to explore the sign value of $\prod_{i=0}^nr_i=(-1)^na_{0}$ in order to define properties of the roots in terms of whether $n$ is odd or even. Is that something I could legitimately do?
The formulas are exact. However, you must be aware that the roots $r_i$ are in general not real, but complex numbers. The subset of real roots may even be empty.
The reason is that if you have the $n$ roots $r_i$ (some of them possibly multiple roots) of your polynomial $P(x)$ of degree $n$, then $$P(x) = a_n\prod_{i=1}^n (x-r_i)$$ Multiplying shows that $a_{n-1} = -a_n\sum_{i=1}^nr_i$ and $a_0 = a_n(-1)^n\prod_{i=1}^n r_i$.