Solving Higher-Degree Polynomials by Synthetic Division and the Rational Roots Test
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- Опубликовано: 26 дек 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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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 RUclips! Cheers, Dave!
i love u. thank you sir
profil fotoğrafındaki yer 3. köprü mü reis
@@muffy4671 3 değil. 2 olması muhtemel
Anyone from 2nd year engineering 😅
University?
Am here dear😂😂
Yes
@@ElinorMerijan NUST
🙋♀️🙋♀️🙋♀️🙋♀️🙋♀️
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.
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.
Wow! This video made a boring and confusing 2 hour lecture into a simple 10 min entertaining video! Thank God for this video.
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!
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.
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.
coolest channel i ever came across the internet
Really good. That helped me a lot.
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
Thank u for helping me learn an entire unit in a single night :)
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.
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 ?
the savior is in time as always big respect
Ty, my brain grew stronger after that.
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
Excellent! Now a fan of synthetic division and Prof Dave!
I never understood it during highschool days ..and now I completely understood this now
0:53 x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)
you just answered all my questions on polynomials in 10 minutes, and I was looking at my notes for clues for days...THANK YOU SO MUCH! also i was late to my first class that started with them hence why I couldn't understand anything
this is great as highschool review! I have my first uni calc test tomorrow that might require me knowing this
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.
Thank you very much sir🙏🙏🙏 with love and respect from INDIA
Such high -quality lectures! Thanks
I was confused up until comprehension , i managed to get all right answers in like 15 minutes
Loved this video now I will top my engineering classes lol😅😅
Lol
Lola
@@shyamsundersharma449 lele
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At a high school level i learnt alot from this Thank you😭🔥🔥🔥🔥
(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.
Great explanation
Professor Dave Explain hcf process by synthetic division.
Great.... Love from India Sir... Namaste 🙏🙏
Me too sir ...🙏🙏🙏
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
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.
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?
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)
thank uuuuu!!! struggling with this for so long, u explained it so well, like its so easy to understand, u saved my life:))
Thank u so much....
I never knew factorizing 3rd degree and 4th degree polynomials was so easy 😅
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?
really amazing i am not understand before clearly that concept
im in alg 2 your videos have saved me man
Wow, even Basic Calculus isn't as tiring as finding a rational root. There are so many options! 😫
Thank you! I was looking for a good vid to share with my students on synthetic division. Awesome!
At 7:06 can't we just solve the quadratic equation with our traditional method of finding roots?
Man! it was live saving!
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
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
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.
Where did find 40 solutions
You got me with the opening jingle!
How can i find out the imaginary roots of a 4th degree..... Equation.
are you from resonance academy?
@@takeallquintillion yea
Tip -we can check solution faster by direct putting possible solution in equation.If equation satisfies than we can continue with division method .
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. :)
I think you can use just the prime factors, in this case: 1, 2 / 1, 3, 5. maybe I'm wrong, but I did this way and got the correct answer :)
@@jeniferleal2004 no as the rule says, you need to include all factors, non prime included
Excellent explanation
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?
To test if x-1 as factor, add all coefficients to see if it equals 0 and add alternative
Have my test today
Wow, this is so clear sir. Awesome, you just gain a new subscriber.
Dude, you are better than #crashcourse .
Excellent explanation .
Thank you so very much .
x^6 + 1 can be factored over the complex numbers, and other fields. You should have told what field you're factoring over.
Thanks for explanation.
dave why cant i just directly plug in the ans that i got from the rational root test ans see which one works
He explained me in a very easy way Thanks
Very good explanation sir
🤩🤩Amazing person sir, you are. This video is very useful for us.
The explanation was lovely
I have forgotten it ... long ago .. and when need to use couldn't recall.. but ur video help alot ... obliged
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.
THANK YOU PROFESSOR DAVE
I am only 9 th grade and yet i can understand and this is because of ur teaching . Thank you so much
Superb video
I just want math2 to be as easy as that in Turkey,unfortunately that is impossible. :( love ur videos btw great work!
Very effective method to solve higher degree polynomials..... ☺☺
That was really good , I like it
I got the same solution but in a different order: (x+1) (x+2) (3x-4) (15x+6). is it still being okey ?
dave you are a godsend
Thank you for helping me in this sum
Lots of Love from India
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
x^6+1 is not prime. It has factors. They are (x^2+1) and(x^4-x^2+1)
4:06 where did x²-5 come from?
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.
THANKS A LOT DUDE...U LITERALLY SAVED MY DAY
Best intro video
Now these methodes i didnt know.
Intro was nice 😁
Amazing Video as always
thanks prof dave its the video i was looking for
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.
Ciao prof ho una dommanda: 6:49 wy (x-1) and not (x+1) ?
The reason why it’s (x - 1) and not (x + 1), is because the factors themselves should equal to zero, so if x = +1, then it should be (x - 1), so it becomes (1 - 1) to give 0
@@trhll5635 Sorry, still not sure in oposite version, and so if x=-1, then it should be (x+1), so it becomes (-1+1) to give 0
@@cezaragalliu347 Correct, if x = -1, then the factor is indeed (x + 1), because the factors should equate to zero
I have been searching for how to find the possible roots. This video explain it. Very helpful.
Wow, your teaching is awesome💖🤩 tysm🌠✨
Is the equation: x^6 + 1 one of the ways we know that there are infinite prime numbers?
what if we don't have a constant term? Say for example, y(4) + y(3) + y(2) = 0
Thank you so much .. sir .. for teaching us like this .. ❤️❤️