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Homework Help: Polynomial factor help

  1. Dec 12, 2008 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    A polynomial [tex]P(x)=(x-b)^7Q(x)[/tex]
    a) Show that [tex]P(b)=P ' (b)=0[/tex]
    b) Hence find a and b, if [tex](x-1)^7[/tex] is a factor of: [tex]P(x)=x^7+3x^6+ax^5+x^4+3x^3+bx^2-x-1[/tex]

    2. Relevant equations
    If [tex]P(x)=Q(x)R(x)[/tex]
    Then [tex]P ' (x)=Q ' (x)R(x)+Q(x)R ' (x)[/tex]

    I can't think of anything for the factoring aspect of the question.

    3. The attempt at a solution
    For a)
    [tex]P(b)=(b-b)^7Q(b)=0[/tex]
    [tex]P ' (x)=7(x-b)^6Q(x)+(x-b)^7Q'(x)[/tex]
    [tex]P'(b)=7(b-b)^6Q(b)+(b-b)^7Q'(b)=0[/tex]

    But for b) I have no idea how to apply anything from a) to answer the question. Any ideas?
     
  2. jcsd
  3. Dec 12, 2008 #2

    gabbagabbahey

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    Re: Factors

    Well if [itex](x-1)^7[/itex] is a factor, then you can write P(x) in the form: [itex]P(x)=(x-1)^7Q(x)[/itex]....So P(1)=__? and P'(1)=___?

    But what are P(1) and P'(1) for [itex]P(x)=x^7+3x^6+ax^5+x^4+3x^3+bx^2-x-1[/itex]?:wink:
     
  4. Dec 12, 2008 #3
    Re: Factors

    From P(1) and P'(1) you will get a simultaneous equation whereby a+b=? and 5a+2b=?
    :wink:
     
  5. Dec 12, 2008 #4

    Mentallic

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    Re: Factors

    Ahh since [tex]P(1)=P'(1)=0[/tex] and by finding [tex]P(1)=6+a+b[/tex] and [tex]P'(1)=37+5a+2b[/tex] from substituting into the equation, I find a and b through simultaneous equations. Thus, [tex]a=-8\frac{1}{3},b=2\frac{1}{3}[/tex]
    I really hope I can pick these ideas up in the test...
     
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