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