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Simplifying Equations with Polynomial denominators/numerators

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data

    I uploaded a picture of the question because I didn't want to confuse people from the way it looks because it is pretty long.


    2. Relevant equations


    3. The attempt at a solution

    I went from the original equation and took an 'x' out of the first numerator and then got the reciprocal of the x^n and multiplied that through

    1. Original Equation
    2. (((x^n)-8)/((x^2n)+(2x^n)+1)) x ((x^2n)-(4x^n)-5)/(x^n)

    I don't know what I can factor out from this point..
    Please help and thanks
    Last edited: Feb 5, 2012
  2. jcsd
  3. Feb 5, 2012 #2


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    x2n = (xn)2

    That in itself may help you. If not, then let u = xn and substitute that.

    Attached Files:

  4. Feb 5, 2012 #3
    Do you mean x^2n= (x^2)^n

    Also as an example, if I took x^n-8 and did that, would it be:

    and i dont know where to go from there

    and for this one:

    what would i do next?
  5. Feb 5, 2012 #4


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    While that's true, SammyS was right to say
    x2n = (xn)2

    So if you were to make the substitution u = xn,
    xn - 8 would become u - 8.
    With the substitution, it would be
    x2n - 4xn - 5
    = (xn)2 - 4xn - 5
    = u2 - 4u - 5

    Now do the same kind of substitution in the first denominator. Doesn't that expression look familiar now? And can't you do something with u2 - 4u - 5?
  6. Feb 5, 2012 #5
    I substituted all the x^n with y's and ended up with:


    Is that all it can get factored and simplified?
  7. Feb 5, 2012 #6
    Also, I think i will lose points by substituting, what is the way without it?
  8. Feb 5, 2012 #7


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    ((xn ) - 2) ((x2n ) + (2xn ) + 4) = x3n - 8 ≠ xn - 8
  9. Feb 7, 2012 #8
    I redid it and got: (((x^n)-8)((x^n)-5))/(x^n)((x^n)+1)

    Is that correct?

    Sorry, I didn't get what you are trying to say
  10. Feb 7, 2012 #9


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    If you mean
    ... then yes, looks right.
  11. Feb 7, 2012 #10


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    Substituting isn't necessary, but makes things look simpler.

    For example, if you were asked to factorize


    then it isn't very clear what needs to be done, but if you let 2a=b then you have

    [tex]x^2+2bx+b^2[/tex] and this is clearly a perfect square, so you can factorize it as so:

    [tex](x+b)^2[/tex] and then you can substitute back in at the end to obtain

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