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

i love u. thank you sir

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.

Hi

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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

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!

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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

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

hmm i think so, check my algebra playlist!

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. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-N-KXStupwsc.html