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 − 756x + 67259 Roots The 5 roots are: x1 = 653 x2 = x3 = 103 x4 = 103 x5 = 653 What is wrong x2756+67259 is reducible. The correct answer should be as the roots are displayed: (x653)^2*(x103)^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]. 
[QUOTE=R. Gerbicz;558540]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 − 756x + 67259 Roots The 5 roots are: x1 = 653 x2 = x3 = 103 x4 = 103 x5 = 653 What is wrong x2756+67259 is reducible. The correct answer should be as the roots are displayed: (x653)^2*(x103)^3 . But you haven't grouped even correctly the roots in that section. [/QUOTE] 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. [/QUOTE] 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].[/QUOTE] Again, the expression is parsed from left to right, and x/2 is not an integer polynomial, so it shows the error. 
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)=(x505340926559057)^2*(x1050418782005819)^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 
All these examples are working now. Thanks for finding the errors.
Please refresh the page to get the current version. 
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 [url]https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisSnAn.pdf[/url] are true, the Galois group is A[sub]n[/sub] or S[sub]n[/sub], 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 [url]https://www.alpertron.com.ar/POLFACT.HTM[/url] Your polynomial x[SUP]12[/SUP] + 45x[SUP]7[/SUP] − 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) 
I fixed the LLL routine and optimized the Hensel Lifting. Now the factorization of polynomials of degree less than 1000 with small coefficients can be done in seconds.

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