How can I find a root of a polynomial, say,

Daanish Wilcox

Daanish Wilcox

Answered question

2022-03-03

How can I find a root of a polynomial, say, x101+x71, without any approximation methods?

Answer & Explanation

paralovut91

paralovut91

Beginner2022-03-04Added 7 answers

If you mean that you wish to express the roots using only the coefficients of the polynomial, the four basic arithmetic operators and radicals, if the polynomial is a quintic or higher, you cannot in general.
There are, of course, many specific cases when you can. For instance, the polynomial in your question factors into (x2x+1)P99(x) where P99(x) is a lengthy 99-th power polynomial that (as far as I can tell) cannot be factorized any further. The quadratic equation x2x+1=0 can be easily solved yielding the complex roots x=1±32. But the one real root of your polynomial, at x=0.980097, cannot be found this way.
Having said that, once you reduced a more complex problem to finding the roots of a rational polynomial, it is often viewed as good as being solved, with the roots having been found for all practical intents and purposes. That is because the values of the roots can always be calculated easily and efficiently. In a sense it's no different from finding the root of the equation x22=0; we cannot do that using the four arithmetic operators only, so we invent the radical symbol which allows us to write x=±2.
Chettaf04

Chettaf04

Beginner2022-03-05Added 7 answers

Applying the formula for real root of an odd degree polonomial , say, x101+x7=1,
The general solution of real root for a polonomial
Xn+xm=1, where (n and m) are odd positive integers and (n is greater than m) x=[11n+2mn+12!n2(3m2n+1)3mn+13!n3+(4m3n+1)(4m2n+1)4mn+14!n4
(5m4n+1)(5m3n+1)(5m2n+1)4mn+14!n4+]
You may notice that this series solution is always a convergent series (very similar to sin(t) or cos(t) or ex …etc), which we consider exact values in our calculations
The general term is so obvious that is
(1)ii!ni[Product(j=1,to j=i1){injm+1}
X=11n+[i=2 to i=](1)ii!ni[Product(j=1,to j=i1){injm+1}
Applying this formula to the above mentioned example we get:
X=[11101862!1012791803!1013721732744!1014
651662673685!1015581592603614626!1016
511522533544555567!1017
441452463744485496508!1018
371382393404415426437449!1019
3013123233343453563673783810!10110

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