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Basic Algebra Question

  1. Sep 11, 2006 #1
    Okay, basic question. I'm working a limit problem and trying to factor part of my fraction:

    t3 + 3t2 - 12t + 4

    Isn't there a way to combine like terms or something? I feel embarrassed asking this lol.
     
  2. jcsd
  3. Sep 11, 2006 #2
    whats the original problem & how did you get that polynomial? maybe you dont really have to factor it...
     
  4. Sep 11, 2006 #3

    berkeman

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    One (brute force) way is to write the factored form, multiply it out, and see if you can solve the simultaneous equations:

    [tex]t^3 + 3t^2 - 12t + 4 = (t+a)(t+b)(t+c) + d[/tex]
     
  5. Sep 11, 2006 #4
    t3 + 3t2 - 12t + 4 / t3 - 4t , as the limit of t approaches 2

    I figured out that that I can get my denominator all the day down to t (t + 2)(t - 2), so I figured that there must be a way to pull that factor out of the numerator.
     
  6. Sep 11, 2006 #5

    0rthodontist

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    So you're looking for a factor of either (t+2) or (t-2). Use polynomial division.
     
  7. Sep 11, 2006 #6

    berkeman

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    Oh, it's a limit question. You call it a basic algebra question, but you may need to apply L'Hospital's (sp?) rule. That's beginning differential calculus. Is that what you're studying? I'll try to scare up a wikipedia entry on it....
     
  8. Sep 11, 2006 #7

    berkeman

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  9. Sep 11, 2006 #8

    berkeman

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    Hmm, that's interesting. Maybe you don't need it after all. That's neat how factoring the denomnator gets away from the infinity.... I hadn't seen that before.
     
  10. Sep 11, 2006 #9

    berkeman

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    In fact, once you factor the bottom to get rid of the infinity, can't you answer the limit question without factoring the numerator?
     
  11. Sep 11, 2006 #10
    That's a little ahead of where we are in my class, but thank you for the info ;)
     
  12. Sep 11, 2006 #11

    berkeman

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    Doh! I see why you need to factor the numerator still -- gotta get rid of that t-2 term to avoid the infinity.:blushing:
     
  13. Sep 11, 2006 #12
    I think so.....I'm actually working on it now lol.
     
  14. Sep 12, 2006 #13

    matt grime

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    No. Factoring the bottom does not get rid of anything until you know it cancels with something in the numerator, so factor the numerator. Since we know one of the factors this is easy if we know polynomial division.

    NB, I am assuming that both numerator and denominator vanish at the relevant point otherwise there is no need to do anything to find the limit.
     
  15. Sep 12, 2006 #14

    HallsofIvy

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    As Orthodontist said before: the only reason you want to factor the numerator is to cancel the (x-2) in the denominator so assume that is a factor in the numerator. Just divide the numerator by x-2.
     
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