Solving Higher-Degree Polynomials by Synthetic Division and the Rational Roots Test
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- Опубликовано: 2 окт 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 RUclips! 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
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.
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
Wow! This video made a boring and confusing 2 hour lecture into a simple 10 min entertaining video! Thank God for this video.
I love when I get all answers correct, but there is always that one I forget to simplify. Overall, I'm satisfied.
0:53 x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)
Loved this video now I will top my engineering classes lol😅😅
Lol
Lola
@@shyamsundersharma449 lele
@@siddhantdhyani4541 lulu
@@nc7375 LiLi
Excellent! Now a fan of synthetic division and Prof Dave!
Who are reading the comments while seeing the video. Hit a like
Seeing the video RIP English
How can i find out the imaginary roots of a 4th degree..... Equation.
are you from resonance academy?
@@takeallquintillion yea
Wow, this is so clear sir. Awesome, you just gain a new subscriber.
Great explanation
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 ?
Thank you so much .. sir .. for teaching us like this .. ❤️❤️
Thank u so much....
I never knew factorizing 3rd degree and 4th degree polynomials was so easy 😅
Your explanation alone was very impressive. I like your articulation of words. Great work professor!
Excellent explanation .
Thank you so very much .
Really good. That helped me a lot.
🤩🤩Amazing person sir, you are. This video is very useful for us.
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!
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?
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)
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
x^6 + 1 can be factored over the complex numbers, and other fields. You should have told what field you're factoring over.
Dave sir thanks your explanation is too good and helpful to solve algebra and calculus
I never understood it during highschool days ..and now I completely understood this now
Ty, my brain grew stronger after that.
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
Excellent explanation
dave you are a godsend
Thanks for explanation.
Professor Dave Explain hcf process by synthetic division.
Wow, even Basic Calculus isn't as tiring as finding a rational root. There are so many options! 😫
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.
Really you’re amazing I never understand mathematics like now
Tip -we can check solution faster by direct putting possible solution in equation.If equation satisfies than we can continue with division method .
thanks a lot professor..
well done..
I just want math2 to be as easy as that in Turkey,unfortunately that is impossible. :( love ur videos btw great work!
At a high school level i learnt alot from this Thank you😭🔥🔥🔥🔥
x^6+1 is not prime. It has factors. They are (x^2+1) and(x^4-x^2+1)
Professor dave can you please explain how to do synthetic division by fractions please👍🙏
lol solving polynomial equation is always kind of fun
It is not
It is not
dave why cant i just directly plug in the ans that i got from the rational root test ans see which one works
I am only 9 th grade and yet i can understand and this is because of ur teaching . Thank you so much
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.
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
really amazing i am not understand before clearly that concept
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?
bro i continued to trying to find a solution within the list when i had a second grade polynomial 😩 i spent i don't know how much time before realizing i could have completed the square
this is great as highschool review! I have my first uni calc test tomorrow that might require me knowing this
At 7:06 can't we just solve the quadratic equation with our traditional method of finding roots?
(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.
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
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
Dude, you are better than #crashcourse .
Thank you very much sir🙏🙏🙏 with love and respect from INDIA
God bless your spirit! Thanks for these videos they’re very helpful
Very interesting
Thank You!!!
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 dave
Why we took negative 3 but ,can I take any number I will take 5 ,6,7
I got the same solution but in a different order: (x+1) (x+2) (3x-4) (15x+6). is it still being okey ?
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
Thank you sir for this video, your videos are always helpful for me 🙏🙏🙏👍please keep on making videos like this
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?
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.
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.
Ciao prof ho una dommanda: 6:49 wy (x-1) and not (x+1) ?
I love you sir, your concept is really nice
What are the roots of x^4 -12x^3+34x^2-12x-90=0
😿
@@englishwords24 😂. 👍
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.
How you came to know that for initial polynomial 2 is the estimation and for the next polynomial -3 is the estimate?
Outstanding
Thank you so much
ayay salamat AGAAHA
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.
Someone get this guy a shirt...😂
Man you have made Life easy.
Thank you Math Jesus
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
Ty great sir!
Thank you so much sir ji 🙏🙏🙏😊😊
but isn't x^6+1 divisible by x^2+1 since we can factorize the former the sum of two cubes?
Bhhot hard
After x-2 do it by grouping it is way easier
What if the possible solutions obtained is a not the zeroes of the equation?🤔
Same thought
@@kaushikbarui6953
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.
THANKKK YOUUU SO MUCHHH
Luv you😘
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
wow! this is the best solution!!!!!!!!!!!!!
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.
ruclips.net/video/N-KXStupwsc/видео.html
so it just comes down to guessing...
Pretty much. It's strategic guessing.
Great.... Love from India Sir... Namaste 🙏🙏
Me too sir ...🙏🙏🙏
Is the equation: x^6 + 1 one of the ways we know that there are infinite prime numbers?
Thank you
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.