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Old 2020-10-01, 20:46   #23
R. Gerbicz
 
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"Robert Gerbicz"
Oct 2005
Hungary

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Thanks. More examples:
polynom: x^5 - 1615*x^4 + 861790*x^3 - 174419170*x^2 + 14998420705*x - 465948627343

Irreducible polynomial factors
The 4 factors are:
x − 653
(x − 103)2
x2 − 756⁢x + 67259
Roots
The 5 roots are:
x1 = 653
x2 = x3 = 103
x4 = 103
x5 = 653

What is wrong x2-756+67259 is reducible. The correct answer should be as the roots are displayed:
(x-653)^2*(x-103)^3 . But you haven't grouped even correctly the roots in that section.


For evaluation part:

polynom: (x+1)/x*(x+1)
Your polynomial
x + 1

This is wrong.


polynom: x/2*2
your answer: Polynomial division is not integer

This could be still ok, if you require that all subresults should be in Z[x].

Last fiddled with by R. Gerbicz on 2020-10-01 at 20:47
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Old 2020-10-02, 01:18   #24
alpertron
 
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Aug 2002
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Quote:
Originally Posted by R. Gerbicz View Post
Thanks. More examples:
polynom: x^5 - 1615*x^4 + 861790*x^3 - 174419170*x^2 + 14998420705*x - 465948627343

Irreducible polynomial factors
The 4 factors are:
x − 653
(x − 103)2
x2 − 756⁢x + 67259
Roots
The 5 roots are:
x1 = 653
x2 = x3 = 103
x4 = 103
x5 = 653

What is wrong x2-756+67259 is reducible. The correct answer should be as the roots are displayed:
(x-653)^2*(x-103)^3 . But you haven't grouped even correctly the roots in that section.
I fixed this error. Please refresh the page and retry.

Quote:
For evaluation part:

polynom: (x+1)/x*(x+1)
Your polynomial
x + 1

This is wrong.
Since the application works with integer polynomials, the slash operator performs the division disregarding the remainder.

This means that when you enter (x+1)/x*(x+1), parsing left to right, the program calculates (x+1)/x = 1 (the remainder 1 of the polynomial division is discarded) and then the program multiplies the previous result 1 by x+1, so the result is x+1.

Quote:
polynom: x/2*2
your answer: Polynomial division is not integer

This could be still ok, if you require that all subresults should be in Z[x].
Again, the expression is parsed from left to right, and x/2 is not an integer polynomial, so it shows the error.

Last fiddled with by alpertron on 2020-10-02 at 01:20
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Old 2020-10-02, 16:23   #25
R. Gerbicz
 
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"Robert Gerbicz"
Oct 2005
Hungary

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OK, more inputs:

(3*x^2)/(2*x+1)
Your polynomial
3x2

Clearly wrong. Following your way it would mean that 3*x^2=(2*x+1)*3*x^2+R(x) and in this case the remainder would be 3rd degree? Meaningless.


x^5 - 4161938199135571*x^4 + 6750425908270471236962285227630*x^3 - 5309242550166291213723686988859597734543042314*x^2 + 2016699400841707878060752590276325131391118358776183137438993*x - 295975920745818133920126480489914719398004640082403818556796416884133107491

Irreducible polynomial factors
The polynomial is irreducible

Roots
The 5 roots are:
The quintic equation cannot be expressed with radicands.

Wrong, because p(x)=(x-505340926559057)^2*(x-1050418782005819)^3

Other such polynoms where you find the roots but display the wrong factors and grouping problems:
x^5 - 569676319*x^4 + 105098371422759466*x^3 - 7011576352652045449773950*x^2 + 195375348220798308339573946630661*x - 1952243687462206905824707435700917386803

and

x^5 - 2876590967309*x^4 + 3229067054667107663190970*x^3 - 1768676572192568691887072530348224746*x^2 + 473795689206607977454622626698064450356613773013*x - 49793205560537089066888851381785438595315265096906181547929
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Old 2020-10-02, 19:42   #26
alpertron
 
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Aug 2002
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All these examples are working now. Thanks for finding the errors.

Please refresh the page to get the current version.
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Old 2020-10-04, 21:20   #27
alpertron
 
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When the irreducible polynomial has degree >= 5, now my code performs the factorization of this polynomials with prime modulus up to 100.

If the conditions of Keith Conrad's paper that you can read at https://kconrad.math.uconn.edu/blurb...galoisSnAn.pdf are true, the Galois group is An or Sn, so my application indicates that the roots of the polynomial cannot be expressed as radical expressions along with the conditions found.

Example of new output from https://www.alpertron.com.ar/POLFACT.HTM

Your polynomial
x12 + 45⁢x7 − 23

Irreducible polynomial factors
The polynomial is irreducible

Roots
The 12 roots are:
x1 to x12 : The roots of the polynomial cannot be expressed by radicals. The degrees of the factors of polynomial modulo 7 are 1, 2 and 9 (the Galois group contains a cycle of length 2) and the degrees of the factors of polynomial modulo 17 are 1 and 11 (the Galois group contains a cycle of prime length greater than half the degree of polynomial)

Last fiddled with by alpertron on 2020-10-04 at 21:35
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