*_All I want for Christmas is a complete proof of the Riemann Hypothesis_*

@joryjones6808

5 жыл бұрын

Me too

@samueldevulder

5 жыл бұрын

Ask to Michael Atiyah ;-)

@AttilaAsztalos

5 жыл бұрын

I have discovered a truly remarkable proof of this theorem which this text box is too small to contain... ;)

@helloiamenergyman

5 жыл бұрын

@@AttilaAsztalos where could you fit it?

@gautamdiwan5952

5 жыл бұрын

@@AttilaAsztalos no shit Fermat

it's a square of a binomial inside the first square root :) √(4 + 2√3) - √3 √(1 + 2√3 + 3) - √3 √( (1 + √3)^2 ) - √3 1 + √3 - √3 1

@tashigawayuno7503

5 жыл бұрын

thanks, u saved my life

@MathNerd1729

5 жыл бұрын

Astute observation, Jose. Bravo!

@rosebuster

5 жыл бұрын

I didn't see this comment before writing my own with the same solution. I would have just replied here.

@jsutinbibber9508

5 жыл бұрын

How does the second one transform into the third one?

@SgtSupaman

5 жыл бұрын

@@jsutinbibber9508, (1 + √3) is the square root of (1 + 2√3 + 3). So, he got the square root and squared it. Basically the steps were this: √4 = √2^2 = 2

saying something equals 3.14... is the clickbait of the math community

@rubenvela44

4 жыл бұрын

There are a lot of things that equal to 3.14.

@contasecreta1234

4 жыл бұрын

@@rubenvela44 like 3 + 0.14

@realbignoob1886

3 жыл бұрын

Lmfao

@uRealReels

3 жыл бұрын

whats funny is that the cube root of 3 + cube root of 2 is about e

@joshuascholar3220

3 жыл бұрын

Not so much. Few would be naive enough to think that pi could be algebraic.

6:50 sqrt(2) + sqrt(3) = x x^2 = (sqrt(2) + sqrt(3))^2 x^2 = 5 + sqrt(24) x^2 - 5 = sqrt(24) (x^2 - 5)^2 = 24 x^4 - 10x^2 + 25 = 24 x^4 - 10x^2 + 1 = 0 Only possible integer solutions are +1 or -1. Subbing them in gives -8 for both, not 0. Therefore solution must be irrational. Therefore sqrt(2) + sqrt(3) must be irrational. 8:04 If the constant is equal to 0 then all other terms are divisible by x, so simply factorise x out of the equation and solve this new equation. The constant being 0 simply means one of the factors of x is 0.

@Mathologer

5 жыл бұрын

Very good :)

@Jo_Es_Chess_Channel

5 жыл бұрын

I think you made a mistake in the third line: x^2=5+2sqrt(6) not 5+sqrt(24)

@TheKingartur007

5 жыл бұрын

@@Jo_Es_Chess_Channel 2 * sqrt(6) = sqrt(24) I think it is right.

@Sordorack

5 жыл бұрын

@@Jo_Es_Chess_Channel well, but it is: 2*sqrt(6) = sqrt(2²) * sqrt(6) = sqrt(4*6) = sqrt(24) is it not? ^^

@Jo_Es_Chess_Channel

5 жыл бұрын

Yeah I'm sorry I thought she did 2x the rest and got 24

I am spending Christmas in Germany this year and we've just exchanged Christmas presents (this happens on the 24th in Germany and not on the 25th). Now it's your turn :)

@gr1nder07

5 жыл бұрын

We do the same, always the day before :)

@andreaaristokrates9516

5 жыл бұрын

In what part of Germany are you, because you said "fröhliche Weihnachten", this sounds weird to me, where I live we/most say "frohe..", but I know there people out there saying it the other way.

@miniwizard

5 жыл бұрын

Wishing you a Merry Christmas - and looking forward to more brain stimulation in the new year!

@maxsch.6555

5 жыл бұрын

Frohe Weihnachten 😄

@John77Doe

5 жыл бұрын

Mathologer X =.sqr(2) + sqr(3) is the same X*X -2 * sqr(2) - 1 = 0 and +1. and -1 are not roots of this quadratic equation. Happy Holidays!!! 🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻

3:05 Holy crap. I'm from the United States and have a university degree in mathematics. I mean, it's totally obvious now that I see it, but I don't recall any of my school teachers or professors ever actually pointing out this trick.

@stephenbeck7222

5 жыл бұрын

Cybis Z every high school algebra 2 and precalculus curriculum I’ve seen has the rational root theorem as the primary method of “guessing” roots of large polynomials to attempt to break them down into at least quadratics (at which point the quadratic formula can be used).

@e11eohe11e

5 жыл бұрын

@Cybis Z Same here, senior getting a BS in Math in NC USA, and haven't seen this even once.

@ProfessorPolitics

5 жыл бұрын

USA also. It was never explained explicitly to my math classes either. I vaguely noticed a pattern when I did algebra homework, but this video was like a lightbulb turning on in my brain.

@DavidSharpMSc

5 жыл бұрын

went to school in Scotland and pretty sure this integral root theorem was never taught to us

@ThePharphis

5 жыл бұрын

it is taught in canada in high school

I'm from Poland and we definitely had both the Integer Root Theorem and the Rational Root Theorem in their proper, full statements in High School. In fact it is still taught, since I tutor a few teenagers and have seen their textbooks. We used them only to find roots of polynomials of order higher then 2 however and not the other way around as you do. So this is indeed a very cool trick. Thank you for the video and Merry Christmas!

@PC_Simo

Жыл бұрын

I’m guessing you took Long Maths, then? I’m from Finland, and don’t remember either from high school (I took Intermediate Maths, though); and we do pretty well, in the Pisa comparisons. 🤔 Also: *_*Than._*

@morgoth4486

Жыл бұрын

yeah, schemat Hornera

It's 00:00 in Germany and Santhologer arrived with this cool present :)

@AlexVasiluta

5 жыл бұрын

@Destro so 60 toffee? I envy you.

@shadowbane7401

5 жыл бұрын

How can it be 00:00?

@AlexVasiluta

5 жыл бұрын

@@shadowbane7401 he means midnight

@DDKKAY

4 жыл бұрын

11:59 time for Grand Finale......this is german.😁

@ffggddss

3 жыл бұрын

@Agustin Vargas Toro: And he arrived to the tune of Tannenbaum. Fred

NY, USA ... never heard of these until university. Sad. However, I was recently visited by the Mathologer Ghost of Christmas Yet to Come ... it warbled at me that .999... = 1. I said, "wait, I've already learned my lesson. I agree ... I agree." "No," it said. "There is more yet to come." Merry Christmas to you and Marty and your staff and families. You have brought much joy and learning over the year.

@Mathologer

5 жыл бұрын

:)

@jessehammer123

5 жыл бұрын

J G Really? At Stuy, I learned this sophomore year.

@willgd6666

5 жыл бұрын

In MA, I learned this in high school.

I'm going to a German school and so far nobody showed us such a useful way to solve our tasks. Thank you and Merry Christmas.

Loved your collab with Flammable Maths

3:05 I was taught this method in 2nd year of high school, but note I was attending mathematics-physics profiled class. I’m from Poland.

@TheZMDX

5 жыл бұрын

Moreover we were taught “real solution theorem” but we didn’t really named it and didn’t consider irrational solutions since polynomial equations that we considered always had a rational solution.

@maciejmanna9246

5 жыл бұрын

Same here (also in Poland).

@pszm5085

5 жыл бұрын

I'm in 2nd year of maths-physics in high school, and I'm also from Poland! :)

@adimoon9

5 жыл бұрын

Yep, exactly the same.

@teodorkowalski3873

5 жыл бұрын

Same. Except for we learned it in 1st class.

In high school in Italy we teach rational root theorem for the factorization of polynomials with "Teorema di Ruffini" and "Regola di Ruffini"

@electric7487

4 жыл бұрын

They teach Ruffini's theorems in secondary schools?!

@3mon3y94

3 жыл бұрын

I d

@raffaeledivora9517

3 жыл бұрын

@@electric7487 Yes.

@dlevi67

3 жыл бұрын

@@electric7487 Yep. First year of high school.

Done three years of maths at the uni in Norway. Never heard of the integral root theorem until now. Neat

Learnt the integral root theorem in high school in Botswana. We then used the solution to divide the polynomial and finally use the quadratic equation

@PC_Simo

Жыл бұрын

You must’ve taken Long Maths, then? I don’t remember being taught that in high school, in Finland. Though, I took Intermediate Maths. Maybe I’ve just forgotten about it. Wouldn’t put it past me, to forget a single topic, in 12-14 years. 🤷🏼‍♂️

Thanks for the video and all other videos of this year. Merry Christmas and Happy New Year!

Merry Christmas Mathologer!

I've taught the integral and rational root theorems in high school math classes in the United States. We then reduced polynomials (using discovered rational factors) to find complex roots as well.

As always, for me at least, another wonderful video. What a great Christmas present. May you and yours and the crew that supports you have the most wonderful of Holidays.

What a wonderful Christmas present! Fröhliche Weihnachten to you and your family as well.

I learned the polynomial guessing trick in school, they also said to start with 1 then -1 then the other factors as they were most likely to be picked by examinators. I'm from Ireland

NYC, Stuyvesant High School, Sophomore Honors Algebra 2. Though I learned it as the Rational Roots Theorem.

I really appreciate that I found this video. I was looking for some lessons and videos which tackle finding irrational root because I failed to find real irrational roots on four items on our activity so I was worrying too much.

You’re awesome! Thanks for the maths and the laughs!! I’m from the US and this was my first time learning both theorems 😀

Love your videos as always! Curious, what animation software you use for these equations? They're quite neat!

sqrt(4+2sqrt(3))-sqrt(3) then this can be written as sqrt((sqrt(3))^2+1^2+2(1)(sqrt(3))-sqrt(3) using identity a^2+b^2+2ab=(a+b)^2 so we get sqrt(((sqrt(3)+1)^2)-sqrt(3) cancelling the square roots we get sqrt(3)+1-sqrt(3) hence 1

@royhedine5153

5 жыл бұрын

You are missing critical brackets ... sqrt((sqrt(3))^2+1^2+2(1)(sqrt(3))-sqrt(3) let A=sqrt(3) sqrt((A)^2+1^2+2(1)(A)-A sqrt((A)^2+1+A ????

@AymanSussy

5 жыл бұрын

nice men

WOW! Thank you! Merry Christmas doc.

I very much enjoyed this video. I've had differential calculus and never came across ether of these theorems. Thank you for your charming presentation.

Thank you for your videos! My University degree is Arts orientated, so I haven’t been focusing on maths as much as beforehand. My enthusiasm for mathematics has been rekindled due to your videos. Again, thank you.

@Mathologer

4 жыл бұрын

That’s great :)

My mother gave me 1 present, my father gave me 2, my grandfather 3, my grandmother 4 and at the end im crying bc i only got -1/12 present

@ajfiaewoipaweipefawejpfwoe1049

5 жыл бұрын

Cristalboy lol well played

@stevethecatcouch6532

5 жыл бұрын

It's worse that you're telling us. If you got -1/12 presents, you must have infinitely many relatives you had to buy presents for.

@looney1023

5 жыл бұрын

If you continue in that fashion, does that mean your great great great . . . grandparents are still alive???

@Chubtato

5 жыл бұрын

Go back to numberphile please

@nischalmr5978

5 жыл бұрын

Steve the Cat Couch epic!

I learned this in secondary school, merry Christmas from Belgium and keep going with the good work !

Merry Christmas. Thanx for the vid.

Is √(2) + √(3) a complex number? √(2) + √(3) = x x^2 = [√(2) +√(3)]^2 x^2 = 5 + √(24) x^2 - 5 = √(24) (x^2 - 5)^2 = 24 x^4 - 10x^2 + 25 = 24 x^4 - 10x^2 + 1 = 0 Possible integer solutions: x = +1 or x = -1. Let f(x) = x^4 - 10x^2 + 1 f(1)= f(-1) = -8, not 0. Therefore √(2) + √(3) must be irrational. What if the polynomial has an ending constant = 0? At least 1 of the solutions is x = 0. Algebraically show that √(4 + 2√3) - √3 = 1 Let x=√(4 + 2√3) - √3 x=√(1 + 2√3 + 3) - √3 x=√( (1 + √3)^2 ) - √3 x=1 + √3 - √3 x=1 Find the equation that has √(4 + 2√3) - √3 as a root? Let x=√(4 + 2√3) - √3 √(4 + 2√3) = x + √3 4 + 2√3 = (x+√3)^2 4 + 2√3 = x^2 + x(2√3) + 3 2√3 = x^2 + x(2√3) - 1 2√3 - x(2√3) = x^2 - 1 2√3 (1 - x) = x^2 - 1 12 (1 - x)^2 = (x^2 - 1)^2 12 (x^2 - 2x + 1) = x^4 - 2x^2 + 1 12x^2 - 24x + 12 = x^4 - 2x^2 + 1 0 = x^4 - 14x^2 + 24x - 11 let f(x) = x^4 - 14x^2 + 24x - 11 f(x) has a root of 1

As for getting that root 3 mess into an integer, here: (note I use rx for the square root of x) (1+r3)^2= 1^2+r3^2+2*1*r3= 1+3+2r3= 4+2r3 Thus, r(4+2r3)-r3= (1+r3)-r3= 1.

@KSignalEingang

3 жыл бұрын

Thanks for this! I kept getting stuck with an equation that either spat out nonsense or something useless like "0x = 0".

Thank you for the present ! once again : very much appreciated ! In France, 35 years ago, for polynomial equations of degree > 2, I was told to try small integers : 1, -1, and 2, -2..

Thank you for your videos. As my family and I are cloudy-minded due to so much Christmas Cheer today, your easy to understand video is very much appreciated.

3:03 I was taught this trick at the 9th year of education in school in Russia.

@zhasulanaset6514

3 жыл бұрын

We taught this trick at 10th grade in Kazakhstan

You really teased pi with the thumbnail

@MathNerd1729

5 жыл бұрын

I know, right? By the way, pi is transcendental so it cannot equal sqrt(2)+sqrt(3) because sqrt(2)+sqrt(3) is algebraic [polynomial: x^4-10x^2+1=0].

@mitchmarq428

5 жыл бұрын

it really DOES equal 3.14...

My favourite Mathologer video of all time! (I know there r many other great ones but I simply love that proof at the end)

In a mathematical training session in the Philippines, I learned how to simplify certain nested radicals. So, I actually knew that the second one wasn't irrational and was equal to one before you told me it was. By the way, great video!

From New Jersey; we learned this in the u/v form, but not really the integral form. Also wanted to say that in the beginning, you briefly brought up root(2) being irrational. A while back, a video went viral where Micheal from Vsauce walked through a proof of root(2) being irrational, but he made a super subtle mistake near the end that pissed me off. He used the fact that an even number squared is even to imply that the square root of an even perfect square was even, which is a converse error. It made me wish you were as big as Vsauce, as you're so thorough in your presentation and explanation I doubt you'd have made such a mistake.

@Laufissa

5 жыл бұрын

So you think that there exists an even perfect square with a root that isn't even?

@rema0126

5 жыл бұрын

Lmao

@MathNerd1729

5 жыл бұрын

Look, odd x odd is odd so it cannot be odd. If it is not even either then what is it?

@looney1023

5 жыл бұрын

@@Laufissa Of course not, but the method of proof was wrong and misleading, especially for the mathematics level of people the video was for. A=>B does not mean B=>A

@Laufissa

5 жыл бұрын

@@looney1023 Ah, I see. It's hard to spot these kinds of mistakes when the statements are very simple and you already know the answer.

Dublin, Ireland. I never heard of the Integral Root Theorem.

I'm from California, USA and I learned about the Integral Root Theorem. Though we learned that in order to find all possible solutions we would need to find our "ps and qs". Where p was the set of factors of the constant, and q was the set of factors of the leading coefficient. Then all possible solutions were made by matching all ps and qs together in the form of p/q.

@ffggddss

3 жыл бұрын

That's the Rational Roots Theorem; the Integral Roots Theorem is a special case of it. Fred

You are the best maths channel I have watched till date :)

I learnt this as the Ruffini method, or something like that, I wasn't paying too much attention tbh. I'm from Spain.

In school in northern England, we were just told to guess random integers

@olbluelips

5 жыл бұрын

ryan2698 - Minecraft Not if the leading coefficient divides the constant term

@ryan2-518

5 жыл бұрын

Ol' Bluelips if it divides it to an integer it’s fine. It’s it’s 3/4 you’re screwed

@brianrobinson4825

5 жыл бұрын

Also from Northern England. Also told to try sensible possible integer values. Even the current A Level exam questions give a root to divide by first. Why was I not told?

👏👏👏👏What took you so long? I have been missing this piece of knowledge since ever! 😁 What a cool video! Thank you very very much!

@Mathologer

5 жыл бұрын

Just wait, three, four videos from now I'll show how a closely related simple result can be used to solve some really ancient problems :)

Burkard, your t-shirt is absolutely brilliant... yet again! 👏

I am from Germany. Never heared about the trick, because it would not work, because while equations are checked to see that getting the solution does not get too overly complicated, the solutions will generally not be nice. We do get fractions, roots and stuff.

@narutosaga12

5 жыл бұрын

kurtilein3 a

@xario2007

5 жыл бұрын

@kurtlein3: Wut? I'm also from Germany and for this kind of problem, where you had to guess the first root to then be able to use pq or abc formula, the first root would usually be an integer or 1/2.

3:03 We were taught the Ruffini method in high school, that was in Spain.

@frabol02

5 жыл бұрын

I was taught the same in Italy

@willa4you

5 жыл бұрын

Me too. Italy. :)

@overfield18

5 жыл бұрын

i just was looking for this comentary, im from spain too

Thanks for the great videos. Merry Christmas. Bon Noël.

I'm from The Netherlands. I have no complaints about the high school math I had, but we did not hear about these integral and rational root theorems. So I know now! Thank you. From all the answers we could build a 'root theorem proliferation map'. :)

Just tell me where do you buy those T shirts! Please! I want to buy all of them!

I learned not exactly this trick for solving polynomial equations, but I cooked up something like that myself without help from textbooks or teacher when I was in high school back in Russia, and was a lot faster with eliminating potential roots once I got a hang of it... And I wasn't even always searching for intereger solution, because if coefficient before the largest power isn't 1, then I would use it as a potential denominator for possible rational fraction solutions too... But I was always good at spotting patterns, I came up with a recursive solution for Tower of Hanoi when I was 6 years old, all on my own, it was a bit cruel of my dad to ask me find the smallest number of moves for 8 tiles high tower of Hanoi just so that I don't bother him, but hey spending 3 hours as a pre-school kid on such a puzzle taught me a lot, and it was a perfect solution too, I still use same thinking I did back then whenever I solve this puzzle...

@trueriver1950

Жыл бұрын

Just think how neat it would've been if you'd already worked out the general result for the Hanoi puzzle and replied instantly "255; and by the way for a stack of ten it's 1023".

@TrimutiusToo

Жыл бұрын

@@trueriver1950 well i mean I was 6 I just learned how to solve it back then, forget general result i didn't even know the multiplication table yet back then...

Here in Brazil we don't learn these root theorems at school. I always thought I should guess integer roots until reducing my polynomial to a second degree polynomial by using long division haha. Thanks for sharing!

Thank you for this Christmas present!

I did learn it in school! Im from iraq xD

@ekuchsbdyshxdhzu7dhdhx949

5 жыл бұрын

I want to say an edgy joke so badly

@yaseralaa9902

5 жыл бұрын

@@ekuchsbdyshxdhzu7dhdhx949 My Christmas present for you is letting you say that joke, go ahead!

@nevokrien95

5 жыл бұрын

so i guess thats why in Israel we dont do cubics

@johnmc1011

4 жыл бұрын

yaser alaa I’m from Iraq and I learn everything from KZread 😂😂😂😂

@hybmnzz2658

3 жыл бұрын

@@ekuchsbdyshxdhzu7dhdhx949 puss

Learned it in grade 7 (at the age of 12) along with Horner's method, polynomial division, factorization, etc. I am originally from Russia.

@RapGeneral11

5 жыл бұрын

Hail slavs? I am form Bulgaria, got taught the same!

I'm in the US, and I was taught the rational root theorem in a high school algebra class. I teach the rational root theorem to my students if I'm ever teaching linear algebra (finding eigenvalues) or differential equations (characteristic polynomial for an nth order linear differential equation with constant coefficients). Great video by the way! Loved it!

Very nice video! I had no idea about this at all (from the US). Merry Christmas!

Soviet school taught me to go through integer solutions first. Vieta formula and stuff. Damn you bloody Bolsheviks!

8:35 "So while the root theorems are fantastic tools for determining the irrationality of an algebraic number, they are of no use for transcendental numbers." But every transcendental number is irrational!

@effuah

5 жыл бұрын

But proving that a number is transcendental may be really hard, if you want to determine for example if π²-e is rational you have some work to do.

@dlevi67

5 жыл бұрын

@@robbraman5869 That is (almost) the definition of transcendental...

Excited for your next video!

Whoaa the ending was neat. These videos always put a giant childish grin on my face Merry christmas!

There is a mistake at 8:37 - transcendental numbers do not have to be real, can be complex as well.

@nathanisbored

5 жыл бұрын

the reals are part of the complex numbers, so just imagine another both around the reals that says 'complex'

@SergiuszOlszewski

5 жыл бұрын

Still, the transcendental numbers box should float over the frame of real numbers. Especially given that they can be imaginary only, for example πi is transcendental.

@nathanisbored

5 жыл бұрын

Sergiusz Olszewski that wouldn’t work either, not every real number is transcendental

@SergiuszOlszewski

5 жыл бұрын

I've put up a visual for you. i.ibb.co/7nFs72g/transcendental.png Tho I'm not sure if every real number has to be either algebraic or transcendental. Edit: the same story is with algebraic numbers - can be complex or imaginary as well.

@becomepostal

5 жыл бұрын

Sergiusz Olszewski I think there are algebraic numbers that are not real.

3:05 - Canada here, and that wasn't ever taught to me.

@reinymichel

5 жыл бұрын

I taught Math in Canada for 33 years and I always taught this. So it must be a regional thing, and depends on your teacher.

This one was excellent - nice job guys!

Thank you so much for another fine video...keep up the work.

0:05 When she cheats on you with a guy just because he's richer

@rubico1894

5 жыл бұрын

Well at least it was with a guy and not the neighbors dog

So cool quick and clean - I love it!

Merry Christmas Burkard and everyone! I never heard about the integer or rational root theorem or a similar trick in school (neither did I during my math degree). Greetings from Austria, Thomas

Fröhliche Weihnachten und einen guten Rutsch ins neue Jahr.

This truly is a gift, some mathematics on your channel that I can actually do!

Firstly, thank you for the video und frohe Weihnachten! I'm from Brazil. I was first presented to the rational root theorem (not it's proof) in my late high school years. Nowadays, it's hard to see it being taught. I would like to suggest a topic related to this one (has to be done after minimal polynomials, if recall it right): unnesting nested radicals. If I'm not mistaken, there's an article about it.

Marry Christmas to you what a nice surprise :D

Und schoene Ruetsch ins neue Jahr! I had no idea. That is so cool.

2:59 I learned it in 9th grade (actually in Spain it's "3rd of ESO", ESO standing for Obligatory Secondary Education in Spanish). Our teacher called it the "Ruffinni method".

3:05 I'm a highshchool student in Catalonia, and we learn this in math class at about fifteen y-o while learning Ruffini's method and we call it "Teorema del residu" (Remainder's Theorem in catalan). Love your videos, by the way.

Verryniice! Merry Xmas to you!

I didn't learn this trick explicitly, but I was taught something in High School that I think is related called the Girard Identities, one of them states that the product of all the roots of a polynomial is equal to the constant term divided by the leading coefficient. I'm from Brazil.

I learned the Integral Root Theorem and Rational Root Theorem in my Sophomore year of high school in the USA in my Pre-Calculus class. It was essentially a class on Euler and a few other mathematicians. However most of the year was spent with polynomial equations, deriving them, and using those derivations so construct shapes with ruler/compass, and prove constants like Pi, e, and i.

I am from New York City and I was never shown anything like the integral root theorem (or rational root theorem) in school or uni, but I read about it on my own time 8) merry Christmas!

I don't know IRT before. Thank you very much. It is the basic and most important theory. Well explained

Your video is really good. I hope I found this channel as a student!

It was taught as the “fundamental theorem of algebra” and we do some synthetic division by one of the zeros to simplify the equation to a quadratic and then just use quadratic formula to find the remaining zeros

I have returned after a six month pilgrimage to understand the generalised hypergeometric function... Worth it

3:00 I learned this as the Rational Root Theorem (which is the generalization and easy consequence of the version you presented). It was basically identical to how you presented it; I don't remember the packaging being very different. I'm from the US East Coast.

amazing way of explaining and an extraordinary video I ve faced telling you what in the last year while I was studying discrete math , a very sophisticated equation faced me so I went to you youtube and searched a lot , until I saw you cute video and the days went on today the same problem faced me so immediately and without thinking I headed towards your video it is really valuable and worth to see all the best and my God help you :)

Great video, as usual

In my American Honors Pre-Calc class, we're taught Rational Root Theorem, which is an extension of what you demonstrate at 1:57. What you do is, you let the lead coefficient be q, and the constant be p. If the polynomial has rational roots, those roots must be in the set of the factors of p divided by the factors of q. We're also told that graphing calculators make Rational Root Theorem obsolete. On exams, we're typically allowed to find all but two roots by graphing. Edit: Oh, you do mention RRT.

Amazing as usual.

I grew up in Spain and I was taught the root theorem, in its rational form, when I was 13-15, as a way yo improve the chance of finding the proper roots of polynomials of integer coefficients when dealing with Ruffini's reduction method.

We've been using that trick with integer roots of polynomial all the time at school. It was important. I couldn't forget it if I wanted to. And I'm from Poland. At times we were using the full version of the theorem to find rational roots, but most often it was the simple case with integer ones. Then we'd just divide the polynomial by (x - root).

learnt the trick in first year high school in Switzerland... Happy christmas btw!

I learned of the Integral Root Theorem in pre-calculus/trigonometry class 2nd year in high school. I went to school in the USA. I was part of an accelerated math track from junior high; half the people taking the class were 3rd year.

Great video and super entertaining!!!

We learned this in year twelve maths in NSW in the '90s. Our selective school had this peculiar way of teaching content, where we'd be taught to be canny to the way exams are put together. So we were taught this for the purpose of self checking, or digging ourselves out of a hole, but we were also taught that, since maths exams mark on working, not on solutions, we still needed to know how to work the question in the expected way. End result was a grade of students who were all hyper cynical about exams, but also had a lot of confidence that we understood the maths we were being tested on, and why we were learning it.

Loved it! Thank you

Great and insightful video, as always! Good to know there are at least 3,524 (as of now) happy boys and girls who got a very nice present for Christmas :) I'm confident that I've learnt both versions of the integral root theorem around 9th grade in school, but it must be noted that I was studying in a specialized math class in a very good school in Moscow, Russia. We were trained for math competitions and olympiads, so they couldn't have overlooked such a cool theorem. Not so sure about Gauss's lemma, but it's possible that we covered that one, too. *Solutions* to the video's puzzles and challenges: 1) sqrt(2)+sqrt(3) is irrational and can be proved by double-squaring and applying the theorem. Note that x = sqrt(2)+sqrt(3) x^2 = 5 + 2*sqrt(6) (x^2-5)^2 = 24 x^4 - 10*x^2 + 1 = 0, and x=1, x=-1 are obviously not solutions to this polynomial equation, meaning that x must be irrational. 2) If the constant term in the equation is zero, then we can note that x=0 is automatically a solution to the equation. Assuming that it is not the value we are looking for, we can divide the equation by x (to the minimal power present), and then apply the integral/rational root theorem to the resulting equation. 3) Massaging sqrt(4+2*sqrt(3))-sqrt(3) into 1 is not that hard. Note that we have sqrt(4+2*sqrt(3)) - sqrt(3) = sqrt( (1+sqrt(3))^2 ) - sqrt(3) = (1+sqrt(3)) - sqrt(3) = 1. 4) Coming up with a polynomial that has the number in challenge (3) as a root: Let x = sqrt(4+2*sqrt(3)) - sqrt(3), then note that x+sqrt(3) = sqrt(4+2*sqrt(3)) => (x+sqrt(3))^2 = 4+2*sqrt(3) => x^2 + 3 + 2*sqrt(3)*x = 4+2*sqrt(3) => x^2 - 1 = 2*sqrt(3)*(1-x) => (x^2-1)^2 = 12*(x-1)^2 => x^4 - 2*x^2 + 1 = 12*x^2 - 24*x + 12 => x^4 - 14*x^2 + 24*x - 11 = 0 This finishes our search.

Yay ! another amazing video :-D I learned things again !!!!