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Factoring a 4th degree polynomial

  1. Jul 25, 2006 #1
    I have the equation

    [tex] f''(t) = 0 =\frac{\ -6t-4t^4+2t^5}{(1+2t)^4} [/tex]

    which I factored to

    [tex] 0 =\frac{ (2t)(-3-2t^3+t^4)}{(1+2t)^4} [/tex]

    and set

    [tex] 2t = 0 [/tex] or [tex] t^4-2t^3-3 = 0 [/tex]

    How am I supposed to factor the second possibility for t? Using guess-and-check (:yuck:) with 0, 1, and -1, I found that t is probably a fraction between -1 and 1. And now I'm stuck for real, because no way am I guess-and-checking random fractions.

    Help, please!
     
    Last edited by a moderator: Jul 25, 2006
  2. jcsd
  3. Jul 25, 2006 #2
    sorry, having latex trouble.

    My second step should read 0=[(2t)(-3 - 2t^3 +t^4)]/(1+t^2)^4
    but I don't think it does, and I can't figure out how to fix the code.
    I'm blaming my computer display
     
  4. Jul 25, 2006 #3

    shmoe

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    All rational roots can be found by the rational root theorem, you don't have to check many:

    http://planetmath.org/encyclopedia/RationalRootTheorem.html

    In any case, when you tried to put in t=-1, you should have gotten 0, no? So you can reduce it to a cubic at least.

    hmm, [tex] 0 =\frac{\ (2t)(-3-2t^3+t^4)}{(1+2t)^4} [/tex]

    odd, it's a cut and paste from yours.
     
    Last edited: Jul 25, 2006
  5. Jul 25, 2006 #4

    0rthodontist

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    The latex doesn't work because you have an extra \ after \frac{
     
    Last edited: Jul 25, 2006
  6. Jul 25, 2006 #5
    Ouch. Another good reason not to guess-and-check: my algebra tends to be better than my arithmetic. Oh, ouch.

    Ummm, I looked at the rational root theorem, but I don't know if I'm using it right.
    Does it say I have some root p/q, where my first coefficient (here 1) is divisible by p and my last coefficient (here -3) is divisible by q? So my answer is -1/3??
    OK, apparantly not, because that doesn't give me zero.

    In either case, it looks like any root will require t to be negative, and in this case, t is time...

    But now I'm curious how this rational root theorem works.
     
  7. Jul 25, 2006 #6

    Office_Shredder

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    The rational root theorem:

    Any rational root of a polynomial will be in the form p/q, where p is a factor of the constant term, and q is a factor of the leading coefficient (so in this case, p is +/-3, 1, and q is +/-1).

    So your best bets would be to check 3, -3, 1, and -1
     
  8. Jul 25, 2006 #7

    shmoe

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    That should just put a space in though? I don't think I changed anything in my cut and paste. Not important though.



    mbrmbrg, you have the rational root theorem reversed, if p/q is a root then q divides your lead coefficient, here a 1, and p divides your constant coefficient, here -3. So you must have q=+ or -1 (so any rational root must also be an integer), and p=+/- 1 or 3. So to find all rational roots, just stick -1, 1, -3, and 3 into your equation. These are not guaranteed to be roots, but any rational root *must* be one of these.

    You must have a positive real roots though. You polynomial is -3 at t=0 and greater than 0 when t is large and positive, so there's at least one root in there somewhere.
     
  9. Jul 25, 2006 #8
    OK, got it now--thanks people!
     
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