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Theorem
Let $P$ be a polynomial of degree $n$ with real or complex coefficients:

Let $z_1, \ldots, z_k$ be real or complex roots of $P$, not assumed distinct.

Suppose $a_n$ is nonzero and:

Then:

Equivalently:

Listed explicitly:

Proof
It suffices to consider the case $a_n = 1$:

The proof proceeds by induction.

Let $\map {\Bbb P} n$ be the statement that the identity below holds for all sets $\set {z_1, \ldots, z_n}$.

Basis for the Induction:

$\map {\Bbb P} 1$ holds because $e_1 \paren { \set {z_1} } = z_1$.

Induction Step $\map {\Bbb P} n$ implies $\map {\Bbb P} {n + 1}$:

Assume $\map {\Bbb P} n$ holds and $n \ge 1$.

Let for given values $\set {z_1,\ldots,z_n,z_{n+1} }$:


 * $\displaystyle \map Q x = \paren { x - z_{n+1} } \prod_{k \mathop = 1}^n \paren { x - z_k }$

Expand the right side product above using induction hypothesis $\map {\Bbb P} n$.

Then $\map Q x$ equals $x^{n+1}$ plus terms for $x^{j-1}$, $1 \le j \le n+1$.

If $j = 1$, then one term occurs for $x^{j-1}$:

If $2 \le j \le n+1$, then two terms $T_1$ and $T_2$ occur for $x^{j-1}$:

The coefficient $c$ of $x^{j-1}$ for $2 \le j \le n+1$ is:

To simplify $c$, use recursion identity:

Then:

Statement $\map {\Bbb P} {n + 1}$ is true and the induction is complete.

Set equal the two identities for $\map P x$:

Linear independence of the powers $1, x, x^2, \ldots$ implies polynomial coefficients match left and right.

Then the coefficient $a_{k}$ of $x^k$ on the left matches $\paren {-1}^{n-k} e_{n-k} \paren { \set {z_1,\ldots,z_n} }$ on the right.

Also known as
Viète's Formulas are also known (collectively) as Viète's Theorem or (the) Viète Theorem.

The Latin form of his name (Vieta) is also often seen.

Also see

 * Definition:Elementary Symmetric Function


 * Elementary Symmetric Function/Examples/Monic Polynomial