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Polynomial ring question

  1. Dec 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Show that x^2 +3*x+2 has four zeros in Z(subscript 6),


    2. Relevant equations



    3. The attempt at a solution

    x^2+3*x+2=(x+1)(x+2)=0=>x=2, and x=1

    6*1=6
    3*2=6

    according to the back of my text book , the other two zeroes for x^2+3*x+2 are x=4 and x=5. is it because x=4 and x=5 are nonfactors of 6?
     
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  3. Dec 14, 2007 #2

    Dick

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    The roots of (x+1)(x+2) over the reals are x=(-1) and x=(-2), which are congruent to 4 and 5 mod 6. So the real question is why x=1 and x=2? The problem is that Z(6) has zero divisors, e.g. 2*3=0. So you can't conclude from (x+1)(x+2)=0 that (x+1)=0 or (x+2)=0.
     
  4. Dec 14, 2007 #3
    Sorry, I meant to right x=-2 and x=-1 . how are x=(-1) and x=(-2) congruent to 4 and 5?
     
  5. Dec 14, 2007 #4

    Dick

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    -2 is congruent to 4 and -1 is congruent to 5, because in each case the difference is divisible by 6.
     
  6. Dec 14, 2007 #5
    I see a pattern if you take the difference between 4-(-2) and 5-(-1) the result for both factors are 6. Would you used this approach to find numbers divisble by 6 given the x values you found from x^2+3*x+2
     
  7. Dec 14, 2007 #6

    Dick

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    That only gives you 4 and 5. It doesn't give you 1 and 2, which are also roots. Unless you know a better system, it's safest to check all possible numbers to see if they are roots.
     
  8. Dec 14, 2007 #7
    I know how to find the roots for x=-1 and x=-2 you just used the system you were taught in high school for breaking up one polynomial expression to two or more polynomial expressions. I wasn't completely sure how to find the zeroes for the other two roots.
     
  9. Dec 14, 2007 #8

    Dick

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    Factoring it is not guaranteed to find all roots for the reasons I pointed out in the first post.
     
  10. Dec 14, 2007 #9
    what alternative approach would I used besides factoring to find the zeroes
    ?
     
    Last edited: Dec 14, 2007
  11. Dec 14, 2007 #10

    Dick

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    Offhand, I don't know of a system if the base isn't a prime. I would say you want to also look at zero divisors. Since 2*3=0 and 4*3=0, (x+1)(x+2)=0 could also mean x+1=2, x+2=3 (which gives you x=1) or x+1=3, x+2=4 (which gives you x=2). Since 6 is a small number, you could also just check all possibilities.
     
  12. Dec 14, 2007 #11

    HallsofIvy

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    Since there are only 6 members of [itex]Z_6[/itex], a "brute strength" method would be to put each number into the polynomial and see what happens!
    Obviously [itex]0^2+ 3(0)+ 2= 2[/itex], [itex]1^2+ 3(1)+ 2= 6= 0 (mod 6)[/itex], [itex]2^2+ 3(2)+ 2= 12= 0 (mod 6)[/itex], [itex]3^2+ 3*3+ 2= 20= 2 (mod 6)[/itex], [itex]4^2+ 3(4)+ 2= 30= 0 (mod 6)[/itex], [itex]5^2+ 3(5)+ 2= 42= 0[/itex] (mod 6).
     
  13. Dec 14, 2007 #12

    Dick

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    Hey Halls, I think over here we call it "brute force" instead of "brute strength". :).
     
  14. Dec 15, 2007 #13

    HallsofIvy

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    Well, you have to have "brute strength" in order to use "brute force"! The important thing is being a brute.
     
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