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Finding the correct value of k so a root is a factor of P(x)

  1. Feb 21, 2015 #1
    1. The problem statement, all variables and given/known data
    P(x)=x^24-2kx^6+k^2

    Find all values of k so that (x-(sqrt(3)/2+i/2)) is a factor of P(x)

    2. Relevant equations

    3. The attempt at a solution
    Letting z=sqrt(3)/2+i/2
    mod(z)=1
    arg(z)=π/6

    z^(24) gives e^(24π/6)i=e^(2πi)=1

    z^(6) gives e^(6π/6)i=e^(π)i=-1

    If (x-(sqrt(3)/2+i/2)) is to be a factor then P((sqrt(3)/2+i/2))=0
    Substituting z^(24) and z^(6) into P(x) gives
    0=1-2*-1*k+k^2
    0=k^2+2k+1
    0=(k+1)^2
    k=-1

    Unfortunately when I plug the new equation with the value of k=-1 into wolfram alpha it doesn't give (x-(sqrt(3)/2+i/2)) as a factor.
     
  2. jcsd
  3. Feb 21, 2015 #2

    haruspex

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    Strange... it clearly is a factor.
     
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