4th degree polynomials

Gauss?

Take the anti-derivative of it and find all local maximums/minimums
easy

Ask someone good at math

wtf you smoking?

I mean he's not wrong...

grouping

plugging in numbers and synthetic division

if it's something like:

[math]x^4 + 3x^2 + 2[/math]

you can just treat it like a quadratic

You'll never actually find a root this way.

He is objectively wrong. You know that critical points don't always correspond to local extrema?

Once you actually get into the real world of math, approximation is your best friend

Also the fact that finding the local maxima/minima of an antiderivative requires finding where the derivative = 0, so you're literally back to where you started

Well obviously he/she meant you could devise some other method of finding local extrema and then use that to conclude the other way. There isn't just exactly one way to do things. But even if you could, it wouldn't work.

But obviously, he means finding the local maxima/minima through other methods.

>other methods

Good for a solid chuckle user. Note how many think you're serious!

Are you retarded? It's so fucking easy to show that, for example, x^2 has only the local minimum at 0 using elementary methods.

sciencedirect.com/science/article/pii/S1110256X14000029

ISBN 978-0-88385-783-0 (maa.org/press/books/beyond-the-quadratic-formula)
ISBN 978-0-8176-4836-7 (link.springer.com/book/10.1007/978-0-8176-4849-7)

Yeah, now do it with 2/3x^7 - 18x^3 + 7.256x^2 - [math]\pi[/math]x + 50253

-1/12

Easy peasy, just substitute into the quadratic formula!

Quartic*

...

...

>be in the club
>this guy comes up and slaps your quintic on the ass
wat do

challenge him to a duel

Why are all mathematicians such qts?

Q-qt

synthetic division

who cares

Set y=0

Brainlets.

You use the rational zero test to find a root, and that's if factoring didn't work. Then use synthetic division.

Or just use synthetic division as long as it crosses the x axis

>rational zero test
Rational zero test? Do you mean the rational root theorem, you absolute piece of shit?

Jesus fuck, highschool kiddies that still call theorems "rules" or "tests" should be gassed.

math.stackexchange.com/q/786

Wow I didn't use your better, more """sophisticated""" word for the process. This is why no one likes math fags that think they're better than everyone else.

No. I just know that "test" is highschool language. That is how I know you are underage and need to leave.

man i havent been here in a long time and i forget how autistic yall are LOL

What if I told you they still teach it as a test in college. Bet you feel real stupid now

In college we saw and proved it as the rational root theorem. If your school calls it a "test" then I do feel stupid, as I now feel bad for insulting someone in your dire conditions.

May your country get out of third world cancer soon. Peace.

Go use pythagorean theorem, tfw brainlet.

The wierstrass M test and others are routinely called tests. Root and ratio convergence tests, etc

This user is correct in that it would be a simple yet efficient method to do, yet I don't believe that is fair to tell the other user that he is stupid because his scholar environment uses some words. I understand the terminology may be wrong by your standards but as long as it conveys what he means, I don't see the harm (maybe a friendly correction or something like that). Also, as this user said
It's not uncommon for shit to be called test, maybe so the people understand that so apply, let's say, a theorem you need to do a sequence of steps and get a a result.

The rational root theorem which basically is, you do synthetic division with a little analysis before hand to get the roots (more precisely, by which terms you divide by looking at your poly.)

find a general solution to his sextic

You guess

/thread

symmetry

>You use the rational zero test
Doesn't work for all relevant polynomials.

Next?

>rational zero test

by passing 10th grade algebra

He wasn't blond.

Uh, you don't get to bring friends.