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Factoring problem

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data

    http://alphacapitalist.com/wp-content/uploads/2012/06/factoringproblem.jpg [Broken]


    2. Relevant equations

    /

    3. The attempt at a solution

    I tried factoring this but it seems to me that author has made an error somewhere. Either solution to this problem is not 1/a-1 or there is error in the problem itself. Why? Well i tried putting a=2 and using author's solution i should get 1 but when i put it in the problem get 1.714. Any thoughts?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 19, 2012 #2

    HallsofIvy

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    I have no clue what this is supposed to be. For one thing it is too small and too dark to read well. Also, there is a symbol between the last two terms inside the parentheses that looks like ":". What does that mean?
     
  4. Jun 19, 2012 #3
    ":" means divided.

    Maybe this will help:
    (3/a-1 - 3a^2+3a+3/a^2-1 : a^4 - a/a^3 +1) * a - a^2/3

    And the solution is 1/a-1
     
  5. Jun 19, 2012 #4
    I did it and got [itex]\frac{1}{a-1}[/itex].

    First I would get rid of the divide then remember that [itex] a^3 -1 = (a-1)(a^2 + a + 1)[/itex]

    See how you go from there.
     
  6. Jun 19, 2012 #5
    Where do you see a^3 - 1?
     
  7. Jun 19, 2012 #6

    micromass

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    Why don't you start by showing what you tried to solve the problem??
     
  8. Jun 19, 2012 #7

    Ray Vickson

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    That is incorrect: the answer is 1/(a-1), not (1/a)-1, which is what your written expression means.

    RGV
     
  9. Jun 19, 2012 #8
    [itex]a^4 - a = a(a^3 - 1)[/itex]
     
  10. Jun 20, 2012 #9
    Here is my stab at it:

    ([itex]\frac{3}{a-1} - \frac{3a^2+3a+3}{a^2-1} : \frac{a^4-a}{a^3+1}[/itex]) * [itex]\frac{a-a^2}{3}[/itex] = ([itex]\frac{-3a^2}{a^2-1}[/itex] * [itex]\frac{(a+1)(a^2-a+1)}{a(a-1)(a^2+a+1)}[/itex]) * [itex]\frac{a-a^2}{3}[/itex] = [itex]\frac{-a(a^2-a+1)(a-a^2)}{(a-1)(a-1)(a^2+a+1)}[/itex] = [itex]\frac{a^2(a^2-a+1)}{(a-1)(a^2+a+1)}[/itex] = [itex]\frac{a^4 - a^3 + a^2}{a^3 - 1}[/itex]
     
  11. Jun 20, 2012 #10
    What did you do to get to ([itex]\frac{-3a^2}{a^2-1}[/itex] * [itex]\frac{(a+1)(a^2-a+1)}{a(a-1)(a^2+a+1)}[/itex])?
     
  12. Jun 20, 2012 #11
    OMFG i made such a stuipid mistake. Instead of first dividing i have subtracted and of course i couldnt get a right solution. I have caluclated and gotten a rght solution.
    Thank you all!
     
  13. Jun 21, 2012 #12
    The symbol ":" has been used in 2 different contexts here. Assuming it means ##\divide##, try cancelling out the "a-1".
     
    Last edited by a moderator: May 6, 2017
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