So I've been thinking about this method for finding one root of a polynomial over the real field .
The only assumption is that the polynomial has a root whose nor is strictly greatest among others roots's norms.
Then we proceed to find this maximum sized root, $r_{max}$.
From the Newton's identities we get the formulae to calculate sum of roots's powers, up to any power.
Let $S_p$ be the sum of the roots's $p$-th powers:
$S_p = r_1^{p} + r_2^{p} + ... r_{max}^{p}+ ... r_d^{p}$,
where $d$ is that polynomial's degree.
Consider $\frac{S_p}{S_{p-1}}$, for arbitrarily natural $p$:
$\frac{S_p}{S_{p-1}} = \frac{r_1^{p} + r_2^{p} + ... r_{max}^{p}+ ... r_d^{p}}{r_1^{p-1} + r_2^{p-1} + ... r_{max}^{p-1}+ ... r_d^{p-1}}$
We can reassemble this, first as:
$\frac{S_p}{S_{p-1}} = \frac{r_{max}^p(\frac{r_1^p}{r_{max}^p} + \frac{r_2^p}{r_{max}^p} + ... 1 + ... \frac{r_d^p}{r_{max}^p})}{r_{max}^{p-1}(\frac{r_1^{p-1}}{r_{max}^{p-1}} + \frac{r_2^{p-1}}{r_{max}^{p-1}} + ... 1 + ... \frac{r_d^{p-1}}{r_{max}^{p-1}})}$
Then as:
$\frac{S_p}{S_{p-1}} = \frac{r_{max}^p((\frac{r_1}{r_{max}})^p + (\frac{r_2}{r_{max}})^p + ... 1 + ... (\frac{r_d}{r_{max}})^p)}{r_{max}^{p-1}((\frac{r_1}{r_{max}})^{p-1} + (\frac{r_2}{r_{max}})^{p-1} + ... 1 + ... (\frac{r_d}{r_{max}})^{p-1})}$
Consider $p\to\infty$, then
$\frac{S_p}{S_{p-1}} = \frac{r_{max}^p(0 + 0 + ... 1 + ... 0)}{r_{max}^{p-1}(0 + 0 + ... 1 + ... 0)} = r_{max}$
So we found the root. By only knowing the Newton's Identities, it is coming through that we could find out the largest of the polynomial's roots, quite, easily?
