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Ive added an attachment and highlited the area that I have a question

  1. Aug 2, 2011 #1
    Ive added an attachment and highlited the area that I have a question about. How where they able to resolve the denominator into factors with out dividing the denominator by the H.C.F.?
     

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  3. Aug 2, 2011 #2

    SammyS

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    Re: Factoring.

    Once it's determined that the numerator in factored form is, x(x+4)(x-1), it's easy to see,by using the remainder theorem, that the only one of these that's a factor of the denominator is x-1.

    The denominator is [itex]7x^3-18x^2+6x+5\,.[/itex]

    Split up -18x2 into -7x2 - 11x2

    You get [itex]7x^3-7x^2-11x^2+6x+5\quad\to\quad7x^2(x-1)-11x^2+6x+5 \,.[/itex]

    Similarly write 6x as -11x - 5x & factor as needed two terms at a time.
     
  4. Aug 2, 2011 #3
    Re: Factoring.

    Got it.. thank you.
     
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