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Run through the steps of this division

  1. Jan 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Please could someone run through the steps of this division ((2(x-1)^2) e^-x)/(1-x)



    2. Relevant equations



    3. The attempt at a solution

    is it neccessary to multiply out this bracket?......im quite confused.

    thanks for help
     
  2. jcsd
  3. Jan 14, 2008 #2

    Tom Mattson

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    Not only is it not necessary to expand the product in the numerator, it is undesirable to do so. When you divide algebraic expressions, you want them to be factored. So let me ask you something, how do you simplify the following?

    [tex]\frac{x-1}{1-x}[/tex]
     
  4. Jan 14, 2008 #3
    to be honest im not sure.... I have looked at a similar problem (x+1)/(x-1) and understand that the aim is to cancel the denominator by writing it above and + 2 to make it the numerator

    therefore x - 1 + 2 / x-1

    so i can see how that becomes 1 + 2/(x-1)

    for some reason i do the same with this and im not getting the correct answer.
     
  5. Jan 14, 2008 #4

    Tom Mattson

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    That is not a similar problem. Go back to the problem I asked you about, and factor -1 out of the denominator. What does the quotient simplify to?
     
  6. Jan 14, 2008 #5
    ok so that then becomes (x-1) / (-1)(-1+x),

    which becomes 1/(-1)(1)
    therefore -1 ?
     
  7. Jan 14, 2008 #6

    Tom Mattson

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    Yes. And that's a piece of the problem you posted in the beginning.

    You have:

    [tex]\frac{2(x-1)^2e^{-x}}{1-x}=\frac{2(x-1)(x-1)e^{-x}}{1-x}[/tex]

    Do you see where you can make a reduction, using what you have just figured out?
     
  8. Jan 14, 2008 #7
    ok so the (x-1)(x-1)/1-x reduces to +1?
     
  9. Jan 14, 2008 #8

    Tom Mattson

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    No. How would that happen? You just figured out that (x-1)/(1-x) reduces to -1.
     
  10. Jan 14, 2008 #9
    i gather im looking at the (x-1)(x-1)/ (-1)(-1+x) cancelling to leave -1.

    then = -2e^-x ??
     
  11. Jan 14, 2008 #10

    Tom Mattson

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    How are you getting that? The (-1+x) on the bottom cancels with one of the factors of (x-1) on the top. After the cancellation, there's still one left!

    Nope.
     
  12. Jan 14, 2008 #11
    does it become -2(x-1)e^-x

    (-2x + 2)e^-x

    = -2xe^-x + 2e^-x
     
  13. Jan 14, 2008 #12

    Tom Mattson

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    Yes, that's it.
     
  14. Jan 14, 2008 #13
    Thanks very much. Sorry for dragging you through that.
     
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