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Polynomial Algebra

  1. Feb 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Let f(x) = anxn + an-1xn-1 + .... + a1x + a0 be a polynomial where the coefficients an, an-1, ... , a1, a0 are integers.

    Suppose a0 is a positive power of a prime number p.

    Show that if [itex]\alpha[/itex] is an integer for which f( [itex]\alpha[/itex] ) = 0, [itex]\alpha[/itex] is also a power of p.

    2. Relevant equations



    3. The attempt at a solution

    I substituted [itex]\alpha[/itex] into the equation in the place of x for each term. I also substituted in pn in the place of a0 as this is a positive power of a prime number p (as given in the question). This gave me:

    f([itex]\alpha[/itex]) = an[itex]\alpha[/itex]n + an-1[itex]\alpha[/itex]n-1 + .... + a1[itex]\alpha[/itex] + pn = 0

    I then decided to isolate pn by moving the other terms to the other side of the equation which gave me:

    pn = -{an[itex]\alpha[/itex]n + an-1[itex]\alpha[/itex]n-1 + .... + a1[itex]\alpha[/itex]}

    Is what I have done so far correct? I now have to show from this that [itex]\alpha[/itex] is also a power of p. I'm unsure what the next step is to do that.

    Can anyone help please?
     
    Last edited by a moderator: Feb 20, 2014
  2. jcsd
  3. Feb 19, 2014 #2

    LCKurtz

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    Do you know the rational root theorem?
     
  4. Feb 19, 2014 #3
    You are correct so far. If you factor out alpha, then you will have that alpha divides ##p^n##.
     
  5. Feb 20, 2014 #4

    Thanks for your help. So because alpha divides p^n this means alpha is also a power of p I'm assuming.
     
  6. Feb 20, 2014 #5
    Yep!
     
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