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# 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 $r_{1}r_{2}r_{3}=-\frac{a_{0}}{a_{3}}$.

## Mathematics Subject Classification

12Y05*no label found*

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