Solving Higher-Degree Polynomials by Synthetic Division and the Rational Roots Test
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- เผยแพร่เมื่อ 5 ก.ย. 2024
- 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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Speaking of the relevancy and the quality of the information, THIS IS THE BEST EVER on the entire TH-cam! Cheers, Dave!
i love u. thank you sir
Outstanding explanation on how to do synthetic division! Not that hard, really. Just tedious trying all the possible rational roots! Thanks, 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.
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
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
I love when I get all answers correct, but there is always that one I forget to simplify. Overall, I'm satisfied.
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. :)
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!
Wow! This video made a boring and confusing 2 hour lecture into a simple 10 min entertaining video! Thank God for this video.
Loved this video now I will top my engineering classes lol😅😅
Lol
Lola
@@shyamsundersharma449 lele
@@siddhantdhyani4541 lulu
@@nc7375 LiLi
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.
According to Indian Education System this is to be taught in class 9th
yeah it sucks, i hope they change the curriculum soon
Same here
I’m here in 7th grade
@@memrman8331 preparing for jee probably?
@@maniyarawat9938 in india these are the topics of class 6 and 7
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
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.
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 ?
Really good. That helped me a lot.
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.
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.
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.
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?
Perhaps, that's what it means to be a real mathematician (I gave up after 2 attempts)
Excellent! Now a fan of synthetic division and Prof Dave!
Professor Dave Explain hcf process by synthetic division.
this is great as highschool review! I have my first uni calc test tomorrow that might require me knowing this
Thank you very much sir🙏🙏🙏 with love and respect from INDIA
(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.
Thank u so much....
I never knew factorizing 3rd degree and 4th degree polynomials was so easy 😅
Thank u for helping me learn an entire unit in a single night :)
really amazing i am not understand before clearly that concept
How can i find out the imaginary roots of a 4th degree..... Equation.
are you from resonance academy?
@@takeallquintillion yea
thank uuuuu!!! struggling with this for so long, u explained it so well, like its so easy to understand, u saved my life:))
dave why cant i just directly plug in the ans that i got from the rational root test ans see which one works
Wow, even Basic Calculus isn't as tiring as finding a rational root. There are so many options! 😫
Thank you sir for this video, your videos are always helpful for me 🙏🙏🙏👍please keep on making videos like this
Great.... Love from India Sir... Namaste 🙏🙏
Me too sir ...🙏🙏🙏
Ty, my brain grew stronger after that.
coolest channel i ever came across the internet
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.
I know it's gonna be good just watching that intro, lmao.
@@abdheshparsad1914
Kaha Gaye
Agr
Answer
Nhi Dena hai
Toh saaf mna kr do n yrr
I never understood it during highschool days ..and now I completely understood this now
I just want math2 to be as easy as that in Turkey,unfortunately that is impossible. :( love ur videos btw great work!
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?
x^6+1 can indeed be factored . the factors are (x^2+1)(x^4+1-x^2)
Yes, you're right
But the roots are imaginary
Excellent explanation .
Thank you so very much .
Ciao prof ho una dommanda: 6:49 wy (x-1) and not (x+1) ?
I have been searching for how to find the possible roots. This video explain it. Very helpful.
God bless your spirit! Thanks for these videos they’re very helpful
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
X=-1
K=5
1+5+(-4)-2)=0
6-6=0
Thank you so much .. sir .. for teaching us like this .. ❤️❤️
🤩🤩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?
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.
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 can you please explain how to do synthetic division by fractions please👍🙏
Great explanation
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
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?
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?
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).
@@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
Wow, this is so clear sir. Awesome, you just gain a new subscriber.
I have forgotten it ... long ago .. and when need to use couldn't recall.. but ur video help alot ... obliged
At a high school level i learnt alot from this Thank you😭🔥🔥🔥🔥
You got me with the opening jingle!
I got the same solution but in a different order: (x+1) (x+2) (3x-4) (15x+6). is it still being okey ?
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 "
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.
@@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.
Is the equation: x^6 + 1 one of the ways we know that there are infinite prime numbers?
Man! it was live saving!
x^6 + 1 can be factored over the complex numbers, and other fields. You should have told what field you're factoring over.
Tip -we can check solution faster by direct putting possible solution in equation.If equation satisfies than we can continue with division method .
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.
Dave sir thanks your explanation is too good and helpful to solve algebra and calculus
thank you so much Dave sir needed this explanation so badly
Wow, your teaching is awesome💖🤩 tysm🌠✨
To test if x-1 as factor, add all coefficients to see if it equals 0 and add alternative
Thank you for helping me in this sum
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.
Thanks for explanation.
That was really good , I like it
I don t understand (factors of the constat term)
In other words, suppose the polynomial has the form:
a*x^n + b*x^(n - 1) + .... + k
n is the degree of the polynomial, i.e. the highest exponent on x
a is the coefficient on the first term
b is the coefficient on the second term
k is the final constant term, that doesn't contain any x's. It's just a constant. In other words, it's the y-intercept.
If you had a polynomial greater than degree 10, you'd have to come up with another set of letters to name it, but I'll call it k, so it can apply to most examples you'd likely do.
@@carultchthankss
Very effective method to solve higher degree polynomials..... ☺☺
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 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?
hmm i believe it does! may want to double check
Have my test today
I am only 9 th grade and yet i can understand and this is because of ur teaching . Thank you so much
thanks prof dave its the video i was looking for
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.
really helpful video thankss!!!
Why'd you use (x-2) in particular, how do we know what factors to choose?
I was asking the same, maybe i missed some parts. Thanks to whom will help on this question.
@@paoloberg70 not sure if you figured it out but it's cuz subbing in 2 into the equation gives 0
@@zafnaahmed3352 He probably planned his examples, using the factored form and then expanding. So he knew in advance which solutions to try.
Really you’re amazing I never understand mathematics like now
He explained me in a very easy way Thanks
How can we find the first two factors . Is it a tukka aur anumaan
Anumaan and match it and if it is correct go to it
Very interesting
Doesn't x^6+1 have x^+1 as a factor?
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).
thanks a lot professor..
well done..
what if you can't guess the 'X'??
If you exhaust all possible rational solutions, then you'd have to use the cubic or quartic formula, whichever is applicable. These are higher degree counterparts to the quadratic formula, although they are complicated and often require a detour to the complex numbers to use. If it is a quintic or anything beyond, then there is no such formula, as proven by Galois. You'd probably have to use Newton's method to approximate the roots.
What are the roots of x^4 -12x^3+34x^2-12x-90=0
😿
@@englishwords24 😂. 👍
THANK YOU PROFESSOR DAVE
Please tell me the solution of given equation x^4+2x^3+3x^2+2x+1=0
Given:
x^4 + 2x^3 + 3x^2 + 2x + 1 = 0
Perform a quartic shift to eliminate the 2x^3 term. Define t such that x = t - 1/2, which will eliminate the x^3 term.
(t - 1/2)^4+2*(t - 1/2)^3+3*(t - 1/2)^2+2*(t - 1/2)+1=0
Expand and simplify:
t^4 + 3/2*t^2 + 9/16 = 0
As you can see, all odd-powered values are now eliminated, and we only have even powers of t. Let u = t^2, and rewrite in terms of u:
u^2 + 3/2*u + 9/16 = 0
Which we can use the quadratic formula to solve for u:
u = (-(3/2) +/- sqrt(9/4 - 4*9/16))/2
Both solutions are u = -3/4, since the square root term simplifies to zero.
This means t = +/- i*sqrt(3)/2
And undo the shift, by subtracting 1/2:
x = -1/2 +/- i*sqrt(3)/2
These are two repeated solutions, that are a complex conjugate pair.
lol solving polynomial equation is always kind of fun
It is not
It is not
what if we don't have a constant term? Say for example, y(4) + y(3) + y(2) = 0
Thank you it helped in lock down😪😢