abstract algebra - Vieta's theorem - Mathematics Stack Exchange
Vieta's theorem states that given a polynomial anxn + ⋯ + a1x + a0 the quantities s1 = r1 + r2 + ⋯ s2 = r1r2 + r1r3 + ⋯ etc., where r1, …, rn are the roots of the given polynomial, are given by si = (− 1)ian − i an. So my question is: can we use this to find all the roots of a polynomial?
algebra precalculus - Using Vieta's theorem for cubic equations to ...
You are guaranteed that this is possible by the fundamental theorem of symmetric polynomials, the proof of which even gives an algorithm for doing this, but it's a pain to do by hand (although it's not a bad exercise in algebraic manipulation).
Alternate proof for Vieta's formula (formula for the summing the …
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.
Roots of a Quartic (Vieta's Formulas) - Mathematics Stack Exchange
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
Solving a quadratic using vietas theorem I keep going in circles.
1. I am trying to solve the quadratic equation x2 − 48x + 432 = 0 x 2 − 48 x + 432 = 0 with out directly factoring OR using the quadratic equation. I am going with vieta. So. r + s = 48 r + s = 48. and. rs = 432 r s = 432. r = 12, s = 36 r = 12, s = 36. but that is too much guessing so I wanted to try a new approach.
Symmetric polynomial and vieta's formulas - Mathematics Stack …
The steps: 1) Sum and write as a quotient of two polynomials. 2) Apply the algorithm to rewrite a symmetric polynomial as a polynomial on the elementary symmetric functions. Replace the occurrences of the elementary symmetric polynomials x1 + x2 + x3 = 0 x 1 + x 2 + x 3 = 0, x1x2 + x1x3 + x2x3 = p x 1 x 2 + x 1 x 3 + x 2 x 3 = p, x1x2x3 = − q ...
Finding a quadratic equation from roots (Vieta's formula)
x2 − (sum of roots)x + (product of roots) = 0 x 2 − (sum of roots) x + (product of roots) = 0. (try and prove this yourself!). Alternatively, if the worst comes to the worst, and you forget this, then just expand [x − (1 − 3–√)][x − (1 + 3–√)] = 0. [x − (1 − 3)] [x − …
Sum of roots: Vieta's Formula - Mathematics Stack Exchange
2014年12月22日 · Sum of roots: Vieta's Formula. Ask Question Asked 9 years, 9 months ago. Modified 5 months ago. Viewed 737 ...
Using Vieta's formula to find the sum of the roots for a given cubic ...
2018年10月13日 · Using Vieta's theorem for cubic equations to derive the cubic discriminant. 3.
Finding the Trace and Determinant with Vieta's Formulas
2016年9月26日 · In studying resultants in the context of proving Bezout's theorem in Algebraic Geometry: A Problem Solving Approach (by Garrity et al), I finally understood the proofs, motivation, and reasoning behind Vieta's formulas.