Solving Higher-Degree Polynomials by Synthetic Division and the Rational Roots Test

By now we are experts at solving quadratics by a number of different strategies. But what about cubics? And quartics? And quintics? Seems pretty daunting, but believe it or not there is a reliable method to solve these higher degree polynomials as well. It's a little more time-consuming, but it can be done! Check it out.
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[@ahmetberatsahin001]

Speaking of the relevancy and the quality of the information, THIS IS THE BEST EVER on the entire KZbin! Cheers, Dave!

[@KoladeEMMANUEL]

i love u. thank you sir

[@fredsalter1915]

Outstanding explanation on how to do synthetic division! Not that hard, really. Just tedious trying all the possible rational roots! Thanks, sir!!

[@yusufdadkhah7561]

1:28-2:54(Synthetic division/how synthetic Divisor for polynomials work) 4:26(check if using the solution of x^-5 holds true for the whole equation shown on the bottom.) 4:30(their individual solutions) 5:29-6:02(rational roots test/possible solutions of 2x^3+3x^2-3x-2) 6:09 (rationial possible solutions test) 8:03(check my compression of it and practice finding solutions of polynomials.)

[@girirajvardhan2722]

Thanks man

[@pranavnyavanandi9710]

But how do you assume that initial root or factor? You just take something between 1 and 9 and try it out? Practically, that may not feasible but at least exam wise... Edit: Figured it out. It's the rational roots test. Should have watched more.

[@user-yr3uj6go8i]

I just finished Calculus 1 with an A (your videos saved my grade) and I still don't know how to do this. I hope I finally understand it this time.

[@abdheshparsad1914]

Hi

[@nerd6414]

Heres a quick trick- If all coefficients add to 0, then 1 is a zero

[@nicolasellis7417]

we appreciate this chief

[@aarohansharma4551]

no way!!! is this real?

[@nerd6414]

@@aarohansharma4551 Try it lmao, what do you need? A 56 page thesis?

[@aarohansharma4551]

@@nerd6414 😁ok I'll try

[@ItsMe-ic5oc]

Holy sh***** wtf YOURE A GENIOUS

[@simonqvistgaard2902]

thank you for making this video, i was working on a equation solver in c++, but had no idea how to make a system for solving huge equations like this, without just guessing. Ive implemented this in my program and it works lige a charm

[@bananprzydawka7129]

Can you show how it works, or send source code?

[@palaaaurinkoon]

​@@bananprzydawka7129 I imagine it might go something like this: 1. def parse_tree("A*(x**3) + B*(x**2) - C(x) + d... == 0") -> return parse_tree (in infix) 2. def eval_equation(parse_tree, xVal) -> do DFS traversal to evaluate parse tree -> if eval == 0 return true, else false 3. create a main function: --3a. loop through values for coefficient and constants (separately) from 1...(a||d) -> if ((a%i==0) push it into array a if (d%i==0) push into array d --3b. write a list comprehension/reducer that takes those two arrays, takes a val from Arr1 (a) and Arr2(b) and combines them (a/b) into new array (C) --3c. prepare a result array and loop through (C)...for every x_value of C, ----3c1. if eval_equation(parse_tree, x_value) -> push x_value into results ----3c2.if eval_equation(parse_tree, -(x_value) -> push x_value into results 4. return results Let me know what you think!

[@Canis_Fatalis]

I love when I get all answers correct, but there is always that one I forget to simplify. Overall, I'm satisfied.

[@ikshasenthet37]

According to Indian Education System this is to be taught in class 9th

[@bosniangamer5509]

yeah it sucks, i hope they change the curriculum soon

[@78anurag]

Same here

[@memrman8331]

I’m here in 7th grade

[@maniyarawat9938]

@@memrman8331 preparing for jee probably?

[@aasmamazumdar2750]

@@maniyarawat9938 in india these are the topics of class 6 and 7

[@jane7963]

Your explanation alone was very impressive. I like your articulation of words. Great work professor!

[@georgesadler7830]

Professor Dave, thank you for an excellent analysis of the Synthetic Division and the Rational Roots Test that is used to solve Higher Degree Polynomials.

[@Bruhhn]

0:53 (x^6 + 1) is not prime polynomial. It could be factorised as (x^2 + 1) (x^4 - x^2 +1)

[@samarthtiwari5532]

x^2 +1 and x^4 - x^2 + 1 both only have imaginary factors so they are prime

[@someone-jl4sj]

@@nimishporwal2658 That is not how you check prime polynomials lmao. x^2+1 is prime polynomial but at x=7 it gives 49+1=50 which is not prime

[@Alexutzugold]

Really good. That helped me a lot.

[@supriyodey3811]

Loved this video now I will top my engineering classes lol😅😅

[@devendeajha1750]

Lol

[@shyamsundersharma449]

Lola

[@siddhantdhyani4541]

@@shyamsundersharma449 lele

[@nc7375]

@@siddhantdhyani4541 lulu

[@Introvert930]

@@nc7375 LiLi

[@martinmoraleta7802]

Wow! This video made a boring and confusing 2 hour lecture into a simple 10 min entertaining video! Thank God for this video.

[@krishnabhattacharya3539]

what if I'm not given the second root i.e (x-3) ... is there any process that can let me skip using Cardan's method ?

[@snekback.]

On the test i got (x-1)(x-4/3)(x+2/5)(15x+30) so it's interesting how the answers may vary depending on the first zero we find.

[@rishabhkumar1631]

Professor Dave Explain hcf process by synthetic division.

[@axolotl4323]

but i got like 40 possible factors for polynomial in checking comprehension. you literally need more than an hour to solve that. or am i doing something wrong?

[@fei_073]

Perhaps, that's what it means to be a real mathematician (I gave up after 2 attempts)

[@surajinsha665]

really amazing i am not understand before clearly that concept

[@jamesmccamish3901]

Excellent! Now a fan of synthetic division and Prof Dave!

[@rishukumar8605]

How can i find out the imaginary roots of a 4th degree..... Equation.

[@takeallquintillion]

are you from resonance academy?

[@jackcantorna1168]

@@takeallquintillion yea

[@HeliosBeats]

this is great as highschool review! I have my first uni calc test tomorrow that might require me knowing this

[@AlexTheNab]

been watching the series for some time but it's my first time commenting. really amazing channel with understandable yet detailed explanation of maths. makes you wonder what those teachers out there are doing, taking a year to teach all that and still people don't understand it.

[@VijayKarajgikar]

You don't need to factor any further from the point of when you got only binomials or trinomials. The maximum degree of your factor is 2 in order for you to not do any synthetic or long division. Then you can, using either completing the square, or quadratic formula to find all your zeros.

[@anirban7493]

It's a shame, I totally forgot about this method and came here to look for other ways to solve higher-degree polynomials without using graphical methods, when I got stuck in a quantum mechanics problem where I am supposed to calculate the eigenvalues and eigenstates of a given Hamiltonian of a quantum system. Thanks, Professor Dave for this walk down the memory lane.

[@abhishek2002_]

Thank you very much sir🙏🙏🙏 with love and respect from INDIA

[@earthlyf3rn]

Thank u for helping me learn an entire unit in a single night :)

[@dima_math]

0:53 x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)

[@yadnyeshfirke9483]

Thank u so much.... I never knew factorizing 3rd degree and 4th degree polynomials was so easy 😅

[@JR-47]

Such high -quality lectures! Thanks

[@kawsar1608]

thank uuuuu!!! struggling with this for so long, u explained it so well, like its so easy to understand, u saved my life:))

[@deepikasharma1414]

Thank you sir for this video, your videos are always helpful for me 🙏🙏🙏👍please keep on making videos like this

[@red_dulge]

I know it's gonna be good just watching that intro, lmao.

[@kanchankumari1150]

@@abdheshparsad1914 Kaha Gaye

[@kanchankumari1150]

Agr Answer Nhi Dena hai Toh saaf mna kr do n yrr

[@SEGAClownboss]

This is the one I'm completely stuck on in terms of that test at the end. It's gonna take me way too long to test every single one of the possible solutions, I think. I had to look for the answer this time. But I'm grateful I get the concept at least.

[@ivorscott]

Rational root test for the comprehension question returns 40 possible solutions. 5 constant factors x 4 leading coefficient factors x 2 ( for +- options) 1,2,4,8,16 / 1,3,5,15 = a very long list indeed I tapped out and ended up looking at the answers then proving why it works. :)

[@trendmemez6701]

Ty, my brain grew stronger after that.

[@elostasam]

I don't understand where the 1 comes from when choosing the factors of the constant term for the numerator and the factors for the leading coefficient for the denominator?

[@AbhishekKumar-rm5bx]

coolest channel i ever came across the internet

[@mertcancapkin8198]

I just want math2 to be as easy as that in Turkey,unfortunately that is impossible. :( love ur videos btw great work!

[@errebusaether]

Wow, even Basic Calculus isn't as tiring as finding a rational root. There are so many options! 😫

[@nellamc5833]

God bless your spirit! Thanks for these videos they’re very helpful

[@huzaifaali3969]

(5:21) how are we getting 1,2 a factor when there is only one constant and that is 2. Same for the denominator how are we getting 1,2 when there is just 2x^3.

[@akhileshnaik4606]

At 7:06 can't we just solve the quadratic equation with our traditional method of finding roots?

[@ahnafalnafis]

Man! it was live saving!

[@maximizerboy]

3:04 Multiplying two negative numbers makes a positive. Why are you factorizing two negative numbers when the end result is still a negative number?

[@Jhay_07]

I never understood it during highschool days ..and now I completely understood this now

[@HimanshuSharma-pz5gm]

x^6+1 can indeed be factored . the factors are (x^2+1)(x^4+1-x^2)

[@kaushikbarui6953]

Yes, you're right

[@mudraboinasailaja9314]

But the roots are imaginary

[@advertenciadecristo3798]

Muito bom. Lembrando que existe fórmula para calcular as quatro raízes da equação polinomial do quarto grau, todavia o processo é tão trabalhoso que buscamos outras formas de resolução.

[@mohammedyousuf2787]

Professor dave can you please explain how to do synthetic division by fractions please👍🙏

[@saguftasimran5082]

Thank you so much .. sir .. for teaching us like this .. ❤️❤️

[@Jourvence777]

Great explanation

[@sibusisomdluli1605]

Hi Prof. Dave what would be the steps in facorizing 6x^3+25x^2+3x-4 ?

[@kfjfkeofitorhf9520]

1/3 6/27+++25/9++6++3/3--4 2/9++25/9+1--4= 27/9+1-4= 3+1--4= 4-4=0

[@supremeleader9579]

Is the equation: x^6 + 1 one of the ways we know that there are infinite prime numbers?

[@jayavallarasum4369]

Great.... Love from India Sir... Namaste 🙏🙏

[@gayatridas7367]

Me too sir ...🙏🙏🙏

[@briandaka3779]

given that x^4+kx^2+4x+2 is a polynomial p(x),when p(X)divide by x+3, the remainder is 8 find the value of K and the remainder when p(x) is divided by x-2

[@kfjfkeofitorhf9520]

X=-1 K=5 1+5+(-4)-2)=0 6-6=0

[@lullabymemo8520]

That was really good , I like it

[@ShlokSengupta]

dave why cant i just directly plug in the ans that i got from the rational root test ans see which one works

[@shahwarmukhtar6780]

I have forgotten it ... long ago .. and when need to use couldn't recall.. but ur video help alot ... obliged

[@maheshkurtkoti5241]

Have my test today

[@sunitakumari6062]

🤩🤩Amazing person sir, you are. This video is very useful for us.

[@ankaiahgummadidala1371]

Excellent explanation . Thank you so very much .

[@indhupriyadharshinisakthiv1104]

I have been searching for how to find the possible roots. This video explain it. Very helpful.

[@Dharmendra1096]

another good way to solve higher order eqn is finding if it is reciprocal eqn or not. this gives couple of methods to solve eqn of degree 4,5,6 etc.

[@ohmygodestovaaserepicopapu383]

8:42 got confused on the 4/3, i tried doing synthetic with 15x - 20 and it the last part equaled to zero, but where will the you put the remaining dropped 15 by the synthethic division?

[@SeanBietz]

You got me with the opening jingle!

[@23jeninaangelind35]

To test if x-1 as factor, add all coefficients to see if it equals 0 and add alternative

[@drakomercury7130]

I got the same solution but in a different order: (x+1) (x+2) (3x-4) (15x+6). is it still being okey ?

[@dakk3]

Am i correct that 8:25 equation has 40 possible solutions? How to know the right one without going through it 40 times?

[@MrDeni23n]

What I did was used this method until I got a quadratic equation (15x^2 - 14x - 8), then used the quadratic factoring method.

[@vasanthamalarvijeyashankhe657]

Thank you for helping me in this sum

[@ClumpypooCP]

lol solving polynomial equation is always kind of fun

[@top_floor_boss]

It is not

[@aarohansharma4551]

It is not

[@lldsll7395]

Doesn't x^6+1 have x^+1 as a factor?

[@ernestschoenmakers8181]

Yes, it can be divided by x^2+1 which results into x^4-x^2+1. This 4th power term can be factored out into (x^2-sqrt(3)x+1)(x^2+sqrt(3)x+1).

[@MiucciaYeung]

dave you are a godsend

[@georgemissailidis3160]

But x^6+1 can be factored: x^6+1 = (x^2+1)(x^4-x^2+1), through noting that x^6+1 is the sum of two cubes.

[@needleinabunchofneedle]

Wow, this is so clear sir. Awesome, you just gain a new subscriber.

[@kennethvalbjoern]

x^6 + 1 can be factored over the complex numbers, and other fields. You should have told what field you're factoring over.

[@aditya-vz9kl]

Thanks for explanation.

[@syedsameed1307]

if none of the solution fraction give a solution .... then what to do?? i am having this problem in this question " X^4+X^3-x^2-3X+2=0 "

[@ssrijan03]

Same problem brother, my question is x^4+2x^3-x^2-6x-3=0 , And I think this method works only for the equations which have atleast one integral root.

[@carultch]

​@@ssrijan03 This method only works when there is at least one rational root. There do exist formulas to solve cubics and quartics, but they are complicated, and are seldom taught outside a math degree. I've used the cubic formula before, and it is not that difficult to make sense of, and it is something that probably should be taught in high school math, even just as a simple introduction without testing students on it. The quartic formula by contrast is a lot more challenging to use. You are probably better off using Newton's method to approximate the roots, for any practical application. Or use Wolfram Alpha, which has this formula programmed into it.

[@masthera73]

At a high school level i learnt alot from this Thank you😭🔥🔥🔥🔥

[@ArmanKhan-nr5dj]

How can we find the first two factors . Is it a tukka aur anumaan

[@alphaproductions211]

Anumaan and match it and if it is correct go to it

[@poonamshah8817]

Wow, your teaching is awesome💖🤩 tysm🌠✨

[@sarthakkharade7112]

thank you so much Dave sir needed this explanation so badly

[@vanguardgaming3721]

Dave sir thanks your explanation is too good and helpful to solve algebra and calculus

[@Chuck25o4m]

I understand the possible solutions in the rational root test will not all work but, will that test always produce an exhaustive list of all possible solutions?

[@ProfessorDaveExplains]

hmm i believe it does! may want to double check

[@progJoe]

Really you’re amazing I never understand mathematics like now

[@Greenwood394]

I know this video is very old but I don't understand why he said x^6 + 1 is prime? Cause you can use sum cubes to get (x^2 + 1)(x^4 - x^2 + 1) please can someone explain why it's prime? Or am I not understanding something?

[@mohammedlarbifaradji4711]

he meant by prime ..when it comes to factoring to two binomials you can't factor x^6+1 into two binomials ...contradictory to x^6-1 you can factor it into (x^3-1)(x^3+1).

[@Greenwood394]

@@mohammedlarbifaradji4711 I did some research and it doesn't have to be binomials. I saw other comments asking my same question so maybe it was a mistake on his part

[@ayushbhandarkar8394]

What are the roots of x^4 -12x^3+34x^2-12x-90=0

[@englishwords24]

😿

[@ayushbhandarkar8394]

@@englishwords24 😂. 👍

[@cailynnthomas769]

THANK YOU PROFESSOR DAVE

[@megawar108]

thanks prof dave its the video i was looking for

[@hugoirwanto9905]

Wow this is what i need!

[@mohamadrzgar98]

thanks a lot professor.. well done..

[@cezaragalliu347]

Ciao prof ho una dommanda: 6:49 wy (x-1) and not (x+1) ?

[@mathandphysicsbysaifanali6169]

x^6+1 is not prime. It has factors. They are (x^2+1) and(x^4-x^2+1)

[@mohit8002]

Tip -we can check solution faster by direct putting possible solution in equation.If equation satisfies than we can continue with division method .

[@Tapleoca]

what if we don't have a constant term? Say for example, y(4) + y(3) + y(2) = 0

[@haha37633]

Very interesting

[@sureshbabupula7151]

Superb video

[@thiruveedulabharathi3619]

He explained me in a very easy way Thanks

[@ModernEducationYt]

I am only 9 th grade and yet i can understand and this is because of ur teaching . Thank you so much

[@navodit1465]

Very effective method to solve higher degree polynomials..... ☺☺

[@kevinalzatefranco5008]

Excellent explanation