Vieta's Formulas

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Background

Let $P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0$, where the coefficient of $x^{i}$ is ${a}_i$. As a consequence of the Fundamental Theorem of Algebra, we can also write

$P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$,

where ${r}_i$ are the roots of $P(x)$.

Let ${\sigma}_k$ be the ${}{k}$th symmetric sum.

Statement

Vieta's says that $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$,

for ${}1\le k\le {n}$.

Proof

[needs to be added]