Irrational Roots

[@evaristegalois6282]

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

[@joryjones6808]

Me too

[@samueldevulder]

Ask to Michael Atiyah ;-)

[@AttilaAsztalos]

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

[@helloiamenergyman]

@@AttilaAsztalos where could you fit it?

[@gautamdiwan5952]

@@AttilaAsztalos no shit Fermat

[@sashabell9997]

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]

Very good :)

[@Jo_Es_Chess_Channel]

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

[@TheKingartur007]

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

[@Sordorack]

@@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]

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

[@unalcachofa]

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

[@AbsoluteHarmacist]

thanks, u saved my life

[@MathNerd1729]

Astute observation, Jose. Bravo!

[@rosebuster]

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

[@jsutinbibber9508]

How does the second one transform into the third one?

[@SgtSupaman]

@@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

[@Lurco8]

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]

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]

yeah, schemat Hornera

[@Mathologer]

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]

We do the same, always the day before :)

[@andreaaristokrates9516]

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]

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

[@maxsch.6555]

Frohe Weihnachten 😄

[@John77Doe]

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!!! 🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻

[@endolp2624]

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.

[@xyz.ijk.]

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]

:)

[@jessehammer123]

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

[@willgd6666]

In MA, I learned this in high school.

[@cybisz2883]

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]

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]

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

[@ProfessorPolitics]

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]

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

[@ThePharphis]

it is taught in canada in high school

[@Fustincho]

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

[@AlexVasiluta]

@Destro so 60 toffee? I envy you.

[@shadowbane7401]

How can it be 00:00?

[@AlexVasiluta]

@@shadowbane7401 he means midnight

[@DDKKAY]

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

[@ffggddss]

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

[@karangupta1825]

There's an identity to add to irrationals: m^(1/2) + (m +1)^(1/2) = (4m+2)^(1/2) It was first derived by ramanujan I have generalized it a bit: M^(1/2) + (m+n)^(1/2) = (4m+2n)^(1/2) I am just 13 and I have been watching your videos for a year 💘 Love from India 🇮🇳

[@mimithehotdog7836]

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

[@OlavRH]

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

[@alicetroise5785]

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

[@electric7487]

They teach Ruffini's theorems in secondary schools?!

[@3mon3y94]

I d

[@raffaeledivora9517]

@@electric7487 Yes.

[@dlevi67]

@@electric7487 Yep. First year of high school.

[@Dradan659]

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]

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

[@mehditabti2658]

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]

Cristalboy lol well played

[@stevethecatcouch6532]

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]

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

[@Chubtato]

Go back to numberphile please

[@nischalmr5978]

Steve the Cat Couch epic!

[@jhawk2402]

11:35 Lets start with √(4 + 2√3) only lets say √(4 + 2√3) = √a + √b square both sides 4 + 2√3 = a + b + 2√ab Because of the square roots, we can deduce this system of equations a + b = 4 2√ab = 2√3 Solving this we get a = 1 and b = 3 (no it doesn't matter which order you put them in) so then going back, √(4 + 2√3) = √1 + √3 = 1 + √3 now adding in the -√3 at the end, 1 + √3 - √3 = 1.

[@catholic_zoomer_bro]

Loved your collab with Flammable Maths

[@menukasharma4890]

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]

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]

nice men

[@TheZMDX]

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]

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]

Same here (also in Poland).

[@pszm5085]

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

[@adimoon9]

Yep, exactly the same.

[@teodorkowalski3873]

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

[@franciskatende1566]

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]

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. 🤷🏼‍♂️

[@BigNWide]

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.

[@F.E.Terman]

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'. :)

[@KnakuanaRka]

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]

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

[@johnwelter7081]

Merry Christmas. Thanx for the vid.

[@miap2605]

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]

That’s great :)

[@NikhilKumar-im8ls]

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

[@benjaminbrady2385]

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

[@MathNerd1729]

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!

[@STD282]

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

[@zhasulanaset6514]

We taught this trick at 10th grade in Kazakhstan

[@billrussell3955]

WOW! Thank you! Merry Christmas doc.

[@Henrix1998]

You really teased pi with the thumbnail

[@MathNerd1729]

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].

[@mirmarq429]

it really DOES equal 3.14...

[@jean-micheltorres6925]

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..

[@jpalreis]

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.

[@Israel220500]

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.

[@kurtilein3]

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]

kurtilein3 a

[@xario2007]

@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.

[@benjaminbrady2385]

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

[@aaryannakhat1842]

Merry Christmas Mathologer!

[@eifbkcn]

My favorite thing with quadratics is solving relative to the vertex, in the form h +/- √(-k/a), as follows ax^2+bx+c (ax^2+bx)+c (and you can set c aside for a while, as it is not needed to find h.) x(ax+b) x(x+b/a), and h = -b/2a Note that if ax^2+b=0, than x is always a root. Therefore, in ax^2+bx+c, the y intercept is c. Another thing to note is that regardless of roots, a parabola will always have the shape ax^2, and therefore if h is the line of symmetry, than k is -a(h)^2+c. you can find zero from k by applying ax^2 to k-k=0. Therefore, if x=0, than ax^2=-k x^2=-k/a x=(+/-) √(-k/a) Thus, you can find the root by centering this parabola on the axis of symmetry, getting x=h (+/-)√(k/a)

[@jessehammer123]

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

[@Unknown10446]

(3:00) I'm a 9th grader from India and this is taught as the "factor theorem".

[@evanyao8475]

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

[@lucasamorim8699]

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!

[@yaseralaa9902]

I did learn it in school! Im from iraq xD

[@samsunggalaxyS24-n5j]

I want to say an edgy joke so badly

[@yaseralaa9902]

@@samsunggalaxyS24-n5j My Christmas present for you is letting you say that joke, go ahead!

[@nevokrien95]

so i guess thats why in Israel we dont do cubics

[@johnmc1011]

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

[@thomaskirsch2209]

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.

[@brendanward2991]

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

[@jmochogi]

I once asked my teacher how I would apply these stuff in life and he told me this: One day, you will be walking home from the market and find the equation blocking your way. It will ask you to solve it before proceeding. I never took my teacher seriously but now I know how mathematics is a gatekeeper to many juicy careers. It took me a while but thank God I am now a mathematics educator, trying to help people learn this stuff in the nicest way possible that makes sense to them.

[@johnchessant3012]

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]

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]

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

[@64standardtrickyness]

For those that care, any algebraic expression of algebraic numbers is algebraic. sketch for a field F the dimension of F[q] (as a vector space over F) is finite iff q is the solution to a polynomial equation over F now if q is algebraic over F[r,s] ( or in laymens terms q is the solution to a polynomial equation using not just F but also r,s e.g note \sqrt( \sqrt 2 + \sqrt 3) is algebraic over Q[ \sqrt 2 , \sqrt 3] Q rationals ) Then there is a theorem of linear algebra that says if F \subfield G \subfield H then the dimension of H over F dim[H / F]is equal to the dimension of H over G times the dimension of G over F which tells us finite dim ext of finite dim ext are finite dim ext thus algebraic ext ( F[q] for q algebraic in F) of algebraic ext are finite dim hence any element of such an extension is algebraic.

[@sergeikalashnikov556]

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

[@frabol02]

I was taught the same in Italy

[@willa4you]

Me too. Italy. :)

[@overfield18]

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

[@ethanyap8680]

3:01 Never learnt it in school, but saw it in olympiad math. Also from down under

[@quackers.]

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

[@olbluelips]

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

[@ryan2-518]

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

[@brianrobinson4825]

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?

[@CastorQuinn]

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.

[@looney1023]

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]

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

[@rema0126]

Lmao

[@MathNerd1729]

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

[@looney1023]

@@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]

@@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.

[@dominicstewart-guido7598]

I learned the trick. They called it the “rational root theorem” and told us to divide all factors of the constant term by all factors of the leading coefficient. From New Orleans, USA

[@AlejandroBravo0]

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

[@SHA256HASH]

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

[@supercobra1746]

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

[@lourdjonsalen8465]

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.

[@professorryze3739]

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

[@ginosantillo4655]

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

[@SergiuszOlszewski]

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

[@nathanisbored]

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

[@SergiuszOlszewski]

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]

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

[@SergiuszOlszewski]

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]

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

[@AlessandroDruetto]

I'm from Italy and studied (in high school) about that method by the name of "Ruffini's Rule": for a polynomial of degree n, we know that an eventual rational solution r must have the form p/q; where p is a divisor of the term that multiplies x^0 and q is a divisor of the term that multiplies x^n.

[@TrimutiusToo]

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]

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]

@@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...

[@ПетяТабуреткин-в7т]

3:05 I was never told about this trick at school, but it was easy to find it myself, although I used a different, easier method to prove that it works. By Vieta's formulas the product of roots of a cubic polynomial is equal to the constant term multiplied by -1, but it's also possible to work it out just for the cubics: Let an equation dx^3 + px^2 + qx + r = 0 have roots a, b and c. d(x-a)(x-b)(x-c) = 0 d(x^2-(a+b)x+ab)(x-c) = 0 d(x^2-(a+b)x+ab)(x-c) = 0 d(x^3-(a+b)x^2+abx - cx^2+(ac+bc)x-abc) = 0 d(x^3-(a+b+c)x^2+(ac+bc+ab)x-abc) = 0 dx^3-d(a+b+c)x^2+d(ac+bc+ab)x-abcd = 0 Therefore, r = -abcd => abc = -r/d.

[@deniskhryashchev7245]

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]

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

[@xCorvus7x]

11:02 1 = sqrt(4 + 2*sqrt(3)) - sqrt(3) is equivalent to 1 + sqrt(3) = sqrt(4 + 2*sqrt(3)). Since both sides are strictly positive, this is equivalent to (1 + sqrt(3))^2 = 1 + 2*sqrt(3) + 3 = 4 + 2*sqrt(3) = sqrt(4 + 2*sqrt(3))^2. A polynomial equation with sqrt(4 + 2*sqrt(3)) - sqrt(3) as a root can be derived as follows: x = sqrt(4 + 2*sqrt(3)) - sqrt(3) Therefore: x^2 = 4 + 2*sqrt(3) + 3 - 2*sqrt(3)*sqrt(4 + 2*sqrt(3)) = 7 + 2*sqrt(3)*(1 - sqrt(4 + 2*sqrt(3)) Further (x^2 - 7)^2 = x^4 - 14*x^2 + 49 = 12*(1 - 2*sqrt(4 + 2*sqrt(3)) + 4 + 2*sqrt(3)) = 12*(5 - 2*(sqrt(4 + 2*sqrt(3)) - sqrt(3))) = 60 - 24*(sqrt(4 + 2*sqrt(3)) - sqrt(3)) = 60 - 24*x Thusly, x^4 - 14*x^2 + 24*x - 11 = 0. Or alternatively, (x^4 - 14*x^2 + 49 - 60)^2 = x^8 - 28*x^6 + 174*x^4 + 308*x^2 + 121 = 576*(7 + 2*sqrt(3) - 2*sqrt(3)*sqrt(4 + 2*sqrt(3))) = 576*x^2. And so, x^8 - 28*x^6 + 174*x^4 - 268*x^2 + 121 = 0. Merry Christmas, everyone, or frohes Fest.

[@denelson83]

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

[@reinymichel]

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.

[@ЮрійЯрош-г8ь]

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

[@mehtapuru]

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

[@rubico1894]

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

[@podemosurss8316]

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".

[@DMSG1981]

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

[@snbeast9545]

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.

[@tissuewizardiv5982]

I'm from Minnesota in the US and we learned it like this. P is the coefficient of the highest power term, Q is the constant term. Any integer solution to the equation just be a factor (positive or negative) of Q/P. Here Q is 6 and P is 1.

[@bensolomons4299]

So I am still being schooled in England, and was only taught how to solve higher order polynomials less than a year ago (I’d taught myself before that, but not the point). We were never taught the Integral Root Theorem but the way that we were taught to find the solutions heavily implied it. We were taught to check integers that divided the constant term because the factor theorem suggests that if f(a) = 0, then (x - a) was a factor of the polynomial. If the polynomial had the coefficient 1 in its highest power then a must be an integer because if it wasn’t, starting with x = c/d, you could rewrite it as (xd-c) = 0 and then when we multiplied that out with the other roots you would get a coefficient that wasn’t 1 for your largest term. It isn’t much of a leap from there to see that all solutions must be integers or irrational

[@wassollderscheiss33]

I learned in school in germany, that you "should" try integer numbers. But they didn't tell us they should be divisors of the constant element (thanks for telling me now, 25 years later...). Also they tought us the "p,q"-formula for quadratic equations and "cardanos formula" for cubic ones. We were told there was a too complicated formula for equations of fourth degree and no algebraicly closed formulas for higher degree equations.

[@RedstoneDefender]

As for 3:05, yes, somewhere in middle school/ early high school (USA) I realized they did that a lot (have root solutions be nice integers, or more generally, specifically choose the questions asked for homework and tests), but it was never explicitly taught. Most my friends did too. Once we got to some of the more complex stuff it became a great way to double check your math answers. Right now in university differential equations I currently use this general idea, that the various questions asked must be specifically chosen (or they risk being unsolvable with the knowledge we've had so far, or potentially "completely" unsolvable (e^(-x^2) anyone?), so occasionally I ask: "hmm is the integrating factor of the characteristic equation some number that makes sense, or did I make a mistake with the negative sign again?", most of the time, if I did something wrong, then extra division, exponents, e^(even more things), logarithms all continually pile up making it easy to spot that something is wrong.

[@cphVlwYa]

I had a pretty awesome algebra 2 teacher here in the US. It wasn't technically part of the curriculum, but he added that little "tip" to the end of our lesson on factoring basically saying, "here's a thing that works, this is why it works, and yes its pretty bad for factoring"

[@김지원-m8q]

11:12 solution Square Root = r r[4+2r(3)]-r(3)=x move -r(3) to the right and square both sides. 4+2r(3)=x^2+3+2r(3)x move 4 and 2r(3) to the right. 0=(x^2-1)+2r(3)(x-1) common factor (x-1) final polynomial (x-1)(x+1+2r(3))=0 so we have another solution -1-2r(3) but r[4+2r(3)]-r(3) is positive so we conclude r[4+2r(3)]-r(3)=1

[@__-cx6lg]

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.

[@PC_Simo]

8:30 I’m pretty sure, given the Root Theorems, that, if the constant term is 0; then, any real solutions must be either 0 or irrational 🙂. *EDIT:* In fact; I can prove it: If the constant term is 0, then we can just move it to the right side, which keeps the right side 0. Then, if the polynomial only has [+]-signs, they cannot possibly cancel each other out; which means that each term must, itself, be 0. But, since they have non-0 coefficients, the solution x must be 0. If the polynomial has [-]-signs, we can move those to the right side, resulting in 2 of those special polynomials that only have [+]-signs and that are equal. To solve each of those, we simply set their value to 0. None of this changes the original root. Then, arguing, as before, we show that x = 0. Therefore, any rational solution must be 0. *Q.E.D. 𑀩* 🙂

[@msdmathssousdopamine8630]

I am a math teacher from Switzerland. I learned the trick (3:05) in middle school (11 yo). Nowadays, it is usually taught (with proof and extended to rational root theorem) during the first year of High school (circa 15 yo) (some middle school teachers still teach it without a proof).

[@minipashki]

We learn this (polynomial equation solving) trick at school in Russia. Usually at an optional class for those who interested in math; or in gymnasiums with physical-mathematical bias. And we learn it for: a1*x^n a2*x^(n-1) ... an*x b=0, where possible rational solutions are (devider of b)/(devider of a1)

[@PC_Simo]

I’m from Finland; and I don’t remember being taught the Integer Root Theorem, in middle school (or high school (Intermediate Maths), for that matter); and Finland’s schools are among the best in the world; but we only solved first-degree equations or partial higher-degree equations, like: x³ + 6 = 0, or: 2x² - 8 = 0, in middle school 🇫🇮🤷🏼‍♂️.

[@QDWhite]

I was taught that integer root trick in Canada, but it was never explained as eloquently as this. I was just told to look for common factors between the coefficient of the highest power and the coefficient of the lowest power and then brute force my way throw them until I find (or didn’t find) a root. I didn’t know I was exploiting a fundamental property of polynomials until watching this video.

[@MuffinsAPlenty]

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!

[@lorenzocarisi]

Hi Mathologer, I'm from Italy and I learned that trick when I was 16 studying in Olympic dispenses in order to win a Math school competition

[@user-wf7uf2jp8x]

I am from Connecticut in America and I learned about the rational roots theorem meaning that any factors of the last coefficient divided by any factors of the first coefficient is a possible solution.

[@robinros2595]

Alternative (and I think more beautiful) proof of the rational root theorem: Suppose u/v is a root of some polynomial f(x) with integer coefficients, with u and v coprime integers. Then (vx - u) is a linear factor in f(x). That is: (vx - u)|f(x). For this to be true, there must be a polynomial g(x) with integer coefficients such that (vx - u)g(x) = f(x). The constant term in f(x) can only be obtained by multiplying u with the constant term of g(x), which again is an integer. So u must divide the constant term of f(x). By the same reasoning, the leading term of f(x) can only be obtained by multiplying vx by the leading term of g(x), which also has a integer coefficient. So v must divide the leading coefficient of f(x). (I'm (ab)using the fact that Z[X] is a unique factorization domain here, I know)

[@michaelempeigne3519]

sqrt ( 2 ) = 1.4142....... and sqrt ( 3 ) = 1.732.......... so sqrt 2 + sqrt 3 = 3.146............... which is not equal to pi

[@thorscape3879]

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.

[@GourangaPL]

In Poland kids learn about that integral root theorem somewhere around 9th grade, and then for this example 3 is an integer root so we divide whole polynomial by (x-3) and then use quadratic formula

[@reetasingh1679]

I learnt the guessing trick in India for solving cubics. We used to guess one root, then use it to reduce the cubic to a quadratic, then use the quadratic formula to find the remaining two roots. Also a suggestion: you could make a video which Mathologerises the proofs for the Cardano or Vieta methods to solve general cubics, and the Ferrari method to solve general biquadratics.

[@covfefe18225]

11:54 This is possible for all polynomials with a degree greater than or equal to 5. For degree 5 polynomials, the ones that can't be expressed in radicals, have a Galois Group of S5 or A5. This is because A5 is simple and non-abelian, therefore making S5 an unsolvable group. More examples of unsolvable polynomials include: x^5 + 3x^3 - x + 1 | x^5 - 2x^3 - 2x^2 + 2x + 2 | x^5 + 3x + 3 | x^5 + x^4 + 3x^3 - 3x^2 - 3x + 2 |

[@88Nieznany88]

I actually had rational root theorem in my middle/high school but never thought of using it to prove whether number is irrational or not. The way we were proving that(for example) square root of 2 is irrational was by contradiction. For example lets assume sqrt(2)=a/b, where a,b - integrals. Therefore 2=a^2/b^2 and 2b^2=a^2. And here you can see if you express left and right side as product of prime numbers, there will be even number of 2's on the right side and odd number of 2's on the left side, because there must be even number of 2's in both a^2 and b^2 and we're multiplying leftside by 2. Right now im at university but ita always nice to learn something new

[@iammaxhailme]

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!

[@doctorslash3922]

I am from Russia. I learned this trick as consequence of the Polynomial remainder theorem which in Russian is called теорема Безу (in Bézout's honor). We used it to solve higher degree equations. I don’t remember when I learned it, probably, in 10th grade. In Russia, schooling lasts 11 years.

[@trueriver1950]

In the West I think that Bézout is credited with a related but different result: if I remember rightly he's the guy who showed that the number of roots (including complex numbers) equals the highest power when all the powers are integers.

[@Zodiarche]

3:00 I learned Polynomdivision in school. You try and error with easy integers, like 1,2,3. When you find a correct one, you divide with (x-n), n is that integer. In your example: (x³-5x²+4x+6)/(x-3) = x²-2x-2 => 0 = f(x) = (x²-2x-2)(x-3)

[@snbeast9545]

11:08 x^4 - 14x^2 + 24x - 11 = 0 The possible solutions are (+-)1 and (+-)11, but the only true solution is one. Algebra (with some minor skipped steps): sqrt(4 + 2sqrt(3)) - sqrt(3) = x sqrt(4 + 2sqrt(3)) = x + sqrt(3) 4 + 2sqrt(3) = x^2 + 2xsqrt(3) + 3 2sqrt(3) = x^2 + 2xsqrt(3) - 1 (2sqrt(3))(1 - x) = x^2 - 1 sqrt(12) = -(x^2 - 1)/(x - 1) 12 = (x^4 - 2x^2 + 1)/(x^2 - 2x + 1) 12x^2 - 24x + 12 = x^4 - 2x^2 + 1 x^4 - 14x^2 + 24x - 11 = 0

[@damiankijano2063]

Here in Uruguay we are taught that a polynomial can be decomposed as a result of the following A. (x-a). (X-b) ... (x-n) where a, b, ..., n are roots of the polynomial equation and A some coefficient , if it expands the independent term is A.a.b ... .n; so the roots must divide the independent term. If any root is zero, the independent term is zero then you can divide the polinomyal by x and then try again.

[@aliasalias2131]

just noticed that at 14:00 you can even pull out v^3, so v^3 has to divide the leading coefficient in this case. so +1,-1 qualifies, but not +2,-2