Vieta's theorem. Ask Question Asked 13 years, 2 months ago. Modified 7 years, 8 months ago. Viewed 3k ...

2016年6月21日 · My second attempt was successful, and I think I've found what is the standard proof, namely using the fundamental theorem of algebra to get linear factors, expanding the right side of the equation, setting the terms equal, and deriving the formula from there. This basic idea can be found in detail at The art of problem solving.

Using Vieta's theorem for cubic equations to derive the cubic discriminant. Ask Question Asked 13 years ago.

2016年4月19日 · $\begingroup$ @Travis Good point, that theorem is always worth a quick test unless you notice something like that though. $\endgroup$ – Justin Benfield Commented Apr 19, 2016 at 9:41

$\begingroup$ @RR: No, Vieta's formula follow directly from the factorization of the quadratic and comparing of coefficients, and works even when the underlying field has characteristic $2$, whereas the quadratic formula doesn't work in that case!

The key facts are Vieta's formulas and the fundamental theorem of elementary symmetric polynomials ...

2018年10月13日 · Using Vieta's theorem for cubic equations to derive the cubic discriminant. 3.

2014年12月22日 · Sum of roots: Vieta's Formula. Ask Question Asked 10 years ago. Modified 8 months ago. Viewed 740 times ...

2019年12月2日 · You can find details in Landau's book Number Theory (Part I, Chapter V, Theorem 77, p. 52) or perhaps other standard references. 52) or perhaps other standard references. Proof .

Alternate proof for Vieta's formula (formula for the summing the roots of a polynomial) 7.