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Help polynomial

  1. Apr 1, 2004 #1
    if p(x ) = x^6 + x^4 + x^2 + 1

    show that the solutions of the equation p(x ) = 0 are among the solutions of the equation x^8 - 1 = 0

    hence factorse p(x ) fully over R
     
  2. jcsd
  3. Apr 1, 2004 #2
    Excuse my ignorance but I don't see how p(x) can ever intersect with the X axis? Its minimum point is (0, 1)... :confused:
     
  4. Apr 1, 2004 #3
    this has complexy number i involved~
     
  5. Apr 1, 2004 #4
    answer = [tex](x^2+1)(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1) [/tex] but i dont see how
     
    Last edited: Apr 1, 2004
  6. Apr 1, 2004 #5
    [tex]p_{(x)} = x^6 + x^4 + x^2 + 1 = x^4(x^2 + 1) + x^2 + 1 = (x^2 + 1)(x^4 + 1) = 0[/tex]

    So either (1) [tex]x^4 = -1[/tex] or (2) [tex]x^2 = -1[/tex]

    Now what are the solutions for [tex]x^8 = 1[/tex]?
    [tex]x^4 = \pm 1[/tex]
    If [tex]x^4 = -1[/tex] we see that it will include the two solutions of equation (1). If [tex]x^4 = 1[/tex] then we can say that [tex]x^2 = \pm 1[/tex] and again, if [tex]x^2 = -1[/tex] we see it will include the two solutions of equation (2).
     
  7. Apr 1, 2004 #6
    did you post this in maths section as well?
     
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