# Vieta’s formula

Suppose $P(x)$ is a polynomial    of degree $n$ with roots $r_{1},r_{2},\ldots,r_{n}$ (not necessarily distinct). For $1\leq k\leq n$, define $S_{k}$ by

 $S_{k}=\sum\limits_{1\leq\alpha_{1}<\alpha_{2}<\ldots\alpha_{k}\leq n}r_{\alpha% _{1}}r_{\alpha_{2}}\ldots r_{\alpha_{k}}$

For example,

 $S_{1}=r_{1}+r_{2}+r_{3}+\ldots+r_{n}$
 $S_{2}=r_{1}r_{2}+r_{1}r_{3}+r_{1}r_{4}+r_{2}r_{3}+\ldots+r_{n-1}r_{n}$

Then writing $P(x)$ as

 $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots a_{1}x+a_{0},$

we find that

 $S_{i}=(-1)^{i}\frac{a_{n-i}}{a_{n}}$

For example, if $P(x)$ is a polynomial of degree 1, then $P(x)=a_{1}x+a_{0}$ and clearly $r_{1}=-\frac{a_{0}}{a_{1}}$.

If $P(x)$ is a polynomial of degree 2, then $P(x)=a_{2}x^{2}+a_{1}x+a_{0}$ and $r_{1}+r_{2}=-\frac{a_{1}}{a_{2}}$ and $r_{1}r_{2}=\frac{a_{0}}{a_{2}}$. Notice that both of these formulas can be determined from the quadratic formula.

More intrestingly, if $P(x)=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}$, then $r_{1}+r_{2}+r_{3}=-\frac{a_{2}}{a_{3}}$, $r_{1}r_{2}+r_{2}r_{3}+r_{3}r_{1}=\frac{a_{1}}{a_{3}}$, and $r_{1}r_{2}r_{3}=-\frac{a_{0}}{a_{3}}$.

Title Vieta’s formula VietasFormula 2013-03-22 15:21:55 2013-03-22 15:21:55 neapol1s (9480) neapol1s (9480) 9 neapol1s (9480) Theorem msc 12Y05 PropertiesOfQuadraticEquation