Results 1 to 4 of 4

Math Help - Factoring polynomials over Z/nZ

  1. #1
    Senior Member
    Joined
    Feb 2010
    Posts
    422

    Factoring polynomials over Z/nZ

    The irreducible quadratic n^2-5n+18 factors as (n-6)(n+1) mod 24. How can I determine this algorithmically?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Pim
    Pim is offline
    Member
    Joined
    Dec 2008
    From
    The Netherlands
    Posts
    91
    While I personally never have seen a problem like this, I would solve it as follows.
    Find two numbers, that when added give the "b". (-6,+1), but (-3, -2) also works here.
    Then use long division with your functions to calculate the remainder of n^2-5n+1/((n-6)(n+1))
    Does this answer your question?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,599
    Thanks
    1421
    (n- 6)(n+ 1)= n^2- 5n- 6. since 6+ 18= 24= 0 (mod24), -6= 18 (mod 24) so this is the same as (n- 6)(n+ 1)= n^2- 5n+ 18 (mod 24).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Feb 2010
    Posts
    422

    Thumbs up

    Sorry, I was unclear. I know this is true, I found it basically how HallsOfIvy did: guess to replace -5 with 16 and factor as usual. My question was, if I did not know the factorization of the polynomial, how could I find it algorithmically?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Factoring Polynomials Help
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 13th 2010, 04:47 PM
  2. factoring polynomials
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 20th 2009, 01:05 AM
  3. Replies: 2
    Last Post: August 22nd 2009, 10:57 AM
  4. Factoring Polynomials
    Posted in the Algebra Forum
    Replies: 5
    Last Post: January 10th 2009, 01:03 PM
  5. Replies: 5
    Last Post: November 29th 2005, 03:22 PM

Search Tags


/mathhelpforum @mathhelpforum