Solving Higher-Degree Polynomials by Synthetic Division and the Rational Roots Test
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- Опубліковано 27 вер 2017
- 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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Outstanding explanation on how to do synthetic division! Not that hard, really. Just tedious trying all the possible rational roots! Thanks, sir!!
Speaking of the relevancy and the quality of the information, THIS IS THE BEST EVER on the entire UA-cam! Cheers, Dave!
i love u. thank you sir
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.)
Thanks man
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.
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.
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 ?
Heres a quick trick- If all coefficients add to 0, then 1 is a zero
we appreciate this chief
no way!!! is this real?
@@aarohansharma4551 Try it lmao, what do you need? A 56 page thesis?
@@nerd6414 😁ok I'll try
Holy sh***** wtf YOURE A GENIOUS
thank uuuuu!!! struggling with this for so long, u explained it so well, like its so easy to understand, u saved my life:))
Really good. That helped me a lot.
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.
Hi
Thank you sir for this video, your videos are always helpful for me 🙏🙏🙏👍please keep on making videos like this
I love when I get all answers correct, but there is always that one I forget to simplify. Overall, I'm satisfied.
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.
Excellent explanation .
Thank you so very much .
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
Can you show how it works, or send source code?
@@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!
Your explanation alone was very impressive. I like your articulation of words. Great work professor!
I have been searching for how to find the possible roots. This video explain it. Very helpful.
Wow! This video made a boring and confusing 2 hour lecture into a simple 10 min entertaining video! Thank God for this video.
Excellent! Now a fan of synthetic division and Prof Dave!
thank you so much Dave sir needed this explanation so badly
Dave sir thanks your explanation is too good and helpful to solve algebra and calculus
this is great as highschool review! I have my first uni calc test tomorrow that might require me knowing this
Thank you so much .. sir .. for teaching us like this .. ❤️❤️
really amazing i am not understand before clearly that concept
Is the equation: x^6 + 1 one of the ways we know that there are infinite prime numbers?
Wow, this is so clear sir. Awesome, you just gain a new subscriber.
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?
Professor Dave Explain hcf process by synthetic division.
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.
thanks prof dave its the video i was looking for
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.
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. :)
God bless your spirit! Thanks for these videos they’re very helpful
That was really good , I like it
Professor dave can you please explain how to do synthetic division by fractions please👍🙏
It is really helpful thank you .
Wow, your teaching is awesome💖🤩 tysm🌠✨
thanks a lot professor..
well done..
I have forgotten it ... long ago .. and when need to use couldn't recall.. but ur video help alot ... obliged
🤩🤩Amazing person sir, you are. This video is very useful for us.
At 7:06 can't we just solve the quadratic equation with our traditional method of finding roots?
Thank u for helping me learn an entire unit in a single night :)
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.
I got the same solution but in a different order: (x+1) (x+2) (3x-4) (15x+6). is it still being okey ?
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.
how do we do the simplification of cubic function to a quadratic function? i,e, taking he factor common
(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.
How you came to know that for initial polynomial 2 is the estimation and for the next polynomial -3 is the estimate?
but isn't x^6+1 divisible by x^2+1 since we can factorize the former the sum of two cubes?
Hi, the numbers that you choose to test them with can only be 1 to 9 (including negative numbers)?
Thank you for helping me in this sum
what if we don't have a constant term? Say for example, y(4) + y(3) + y(2) = 0
I am just curious about how the technique was “devised” and its general form.
Thanks for explanation.
thank you so much now I understand everything !
What should we do if there the constant term has decimals i.e.9.125
To test if x-1 as factor, add all coefficients to see if it equals 0 and add alternative
Thank u so much....
I never knew factorizing 3rd degree and 4th degree polynomials was so easy 😅
can i use this for binomial tho?
dave why cant i just directly plug in the ans that i got from the rational root test ans see which one works
idk if it would work but is possible that when we got x-1 at 6:55 we could use quadratic formula on the rest (sry if its dumb question)
He explained me in a very easy way Thanks
0:53 (x^6 + 1) is not prime polynomial. It could be factorised as
(x^2 + 1) (x^4 - x^2 +1)
x^2 +1 and x^4 - x^2 + 1 both only have imaginary factors so they are prime
@@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
really helpful video thankss!!!
But can I have a question @Professor_dave_explains ? how can we obtain the roots of any higher degree-equation even in complex numbers?
Just curious.
There is a master formula for quadratics, cubics, and quartics, but quartics are the end of the line. Galois proved that there can be no quintic formula, or anything beyond, if you are limited to arithmetic, powers, roots, exponentials, logarithms, and trigonometry. In special cases, it is possible to extract the roots of a higher degree polynomial, despite no such formula existing, but only in those special cases. Such as triquadratic 6th degree equations, where you can simply replace x^3 with w, and solve for w. You can also use the rational roots theorem and trial and error, if you are lucky enough to have a rational root.
I can refer you to Mathologer's video on how to make sense of the cubic formula, which I've mastered after exploring on my own with his help. He also reveals the quartic formula at the end, but I've yet to make sense of it.
ua-cam.com/video/N-KXStupwsc/v-deo.html
coolest channel i ever came across the internet
THANK YOU PROFESSOR DAVE
Hi Prof. Dave what would be the steps in facorizing 6x^3+25x^2+3x-4 ?
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
Amazing Video as always
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?
Great explanation
THANKS A LOT DUDE...U LITERALLY SAVED MY DAY
Hello Professor Dave. Have you ever made a Dividing Polynomials video? Can't find one from you, and I'd love to see this concept from your perspective.
hmm i think so, check my algebra playlist!
Can binary search be used since can get an upper lower bound using the rrt test? Also does real number distance from zero give a direction for that search? Math people better than me since it could be used to speed up solving.
Loved this video now I will top my engineering classes lol😅😅
Lol
Lola
@@shyamsundersharma449 lele
@@siddhantdhyani4541 lulu
@@nc7375 LiLi
The explanation was lovely
How will u come to know to take 2 only to that equation tell me anyone
Really you’re amazing I never understand mathematics like now
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.
Great lecture
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?
Very good explanation sir
What if the polynomial is missing something, like x^4 + 3x^2 + x +1, lacking the x^3. Is it just a 0 in place of the x^3 and that is used for the synthetic division?
yep!
Thank you very much sir🙏🙏🙏 with love and respect from INDIA
Ciao prof ho una dommanda: 6:49 wy (x-1) and not (x+1) ?
Excellent explanation
is it okay that results 4/3 and -2/5 I reached by using a quadratic formula?
You got me with the opening jingle!
Very nice explanation...
Can someone please explain how they got the answer for the comprehension check question
I just want math2 to be as easy as that in Turkey,unfortunately that is impossible. :( love ur videos btw great work!
How you find possible solutions to try plzzz tell me...
How could you take some random values as it's factor. For example how to took exactly 2 as the factor for the polynomial equation
He probably set up his examples, by starting from known factors, and then expanding to find the standard form of the polynomial to give as the example.
This means, he knew in advance which values to try. The rational root theorem allows us to narrow down the possibilities, and he chose from those possibilities.
Sir,how to find roots of X^4+10X-56=0
Am i correct that 8:25 equation has 40 possible solutions? How to know the right one without going through it 40 times?
What I did was used this method until I got a quadratic equation (15x^2 - 14x - 8), then used the quadratic factoring method.
Do we always add and multiply when doing synthetic division?
Yeah
Is this possible for qubic equation
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?
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.
you are a hero!! love you man!!
Does the rational root test work for power 4??