Vieta's Formula

This is, by far, the most important polynomial formula I have ever come across. It works for real and complex roots without fail. It basically interprets the expansion of the factors of a polynomial, no matter the degree. this is a must know for any math Olympian

[puepx]

Trivia: Viète (or Vieta in Latin form) was a lawyer by profession, and did not bother to make this Formula publicized since he thought it was obvious or too basic.

[paulcooper]

Wow, I have never heard of Vieta Formula. Excellent presentation as usual.

[baldaiomir]

in my country we study Vieta's formula in middle school, along with discriminant. but as I remember, not in depth, because we only learn the -b/a formula for sum of roots and c/a for multiplication of roots

[kragiharp]

Long forgotten.
Thank you for reminding.
❤️🙏

[Sigma.Infinity]

4:55 I've never seen inequalities under the summation sigma before. I tried to look it up but there is very little online about it. Great video. I enjoy your gentle humour.

[thatcrp]

Lmao, I was taught this in yesterday's class and I couldn't understand it, and here you are!
Thank you very much Sir

[tankenjopatrickhshs]

never been this early man! love your videos! your enthusiasm and knowledge always brighten up my day lol

[pedropiata]

I Saw that Pattern when I tried to expand (x+a)(x+b)(x+c), then I did (x+a)(x+b)(x+c)(x+d), after that, I did (x+a)(x+b)(x+c)(x+d)(x+e)
After analasizing all the results I came to a formula, witch if u turn the other side to 0 and divede both sides by a_n, you get basicly the vieta's formula, I am so proud of myself😁

[panagiotisvlachos]

How can we prove Vieta's Formula?? That would be an interesting video! Greetings from Hellas!

[joyneelrocks]

These are observational formulas so they r called Vieta’s Laws. Also, i think the second equation is wrong. It should be:
(r_1r_2 + r_1r_3 + … + r_1r_n) + (r_2r_3 + r_2r_4 + … + r_2r_n) + … + (r_{n-1}r_n) = a_{n-2} / a_n

[beapaul]

This formula is similar to the formula that we use for finding probability in a binomial distribution.

[Harrykesh]

Super useful theorem in Algebra Problems

[pauljackson]

You should do things on quaternions instead of just complex numbers.

[DukeofEarl]

Way back when in the UK, I did 2 'O' levels and 2 'A' levels in school and higher school maths, plus a lot of maths during an electronic engineering degree and was never taught this!!

[holyshit]

Maybe series about symmetric polynomials
Then Vieta formulas can be put somewhere in such series
Symmetric polynomials are such polynomials that are invariant to permutation of variables

[bessaniozuber]

at any point are you planning to do a bunch of series on various topics from ground up like a 15 ep run on matrices or any

[donsena]

So, is there a way we can use these results to derive the roots of the cubic equation?

[lukaskamin]

In school we only learned them for quadratic equations, I'm curious what is the connection between this and the algorithm for factoring quadratic polynomials, when you guess coefficients that if multiplied, are to be equal to a*c, but their sum is b. Those numbers should be opposite to the roots of a quadratic equation that could be guessed according Vieta's formula? Maybe you have a video, explaining how this algorithm of factorisation is proven, why it works. I feel like it's something to do with Vieta's formula

[redroach]

Could you use this to solve for the roots by setting up a system of equation and solving for a, b, c or does that not work?

[bobkitchin]

Can Vieta Formulas be used to numerically find the roots? It would seem plausible that a computer algorithm could use the formulas to iteratively narrow in on them.