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Factoring problem that has me stumped!

  1. May 26, 2004 #1
    Given:

    [tex] s(t) = 1/2t^3-5t^2+3t+6 [/tex]

    I'm trying to find all values of t where s(t) = -30

    My first thought is to solve for 0 hence:

    [tex] 1/2t^3-5t^2+3t+36=0 [/tex]

    I know the answers are t=4 and t=8.196 but I can't get to it...I'm assuming I need to factor this down but I'm can't see it. Any help/hints would be most appreciated as I've been banging my head against a brick wall for some time now...
     
  2. jcsd
  3. May 26, 2004 #2

    arildno

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    You know that 4 is a root of the polynomial in your last equation.
    Use polynomial division to compete the factoring.
     
  4. May 26, 2004 #3
    As above, but divide (x-4) into 1/2t^3-5t^2+3t+36

    Then when you get the second polynomial use the quadratic solution to get t=8.196
     
  5. May 28, 2004 #4
    Thanks for the hints - reading up on polynomial division (which I wasn't familiar with) I have found that the factors are:

    [tex] (t-4)(1/2t^2-3t-9) [/tex]

    However, (and I realise I'm getting slightly off topic here) how would I even arrive at (t-4) being one of the original factors. Using GCF I can easily see that:

    [tex] (t)(1/2t^2-5t+3)+36 = 0 [/tex]

    But the jump to t-4 has got me stumped!

    Any pointers to threads/websites on factoring polynomials/finding roots of polynomials would be most appreciated.
     
    Last edited: May 28, 2004
  6. May 28, 2004 #5

    arildno

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    I believe Fermat developed various techniques in order to make good, reasoned guesses for the roots of polynomials.

    However, when meeting a polynomial in a textbook that you don't immediately recognize the roots of, remember that the constant term is the product of the roots.

    Therefore, one way of arriving at 4 as a root is to examine the integer factors of 36, and see which of these (if any!) might be a root.
     
  7. May 30, 2004 #6
    putting the equation equal to -30 u can make a cubic equation which will be then easy to solve for
     
  8. May 30, 2004 #7
    Use descarte's rule of signs to find the nature of the roots (positive, negative, or complex).

    After that take X minus all factors (positive and negative) of the coefficient of the highest degree of your variable over the factors of your constant.

    3X^3 + 5X^2 - 2X + 8 =0

    so your possible (real) factors are going to be (3/8), (3/4), (3/2), 3, (1/8), (1/4), (1/2), 1 and their inverses.

    Well, now you're left with 14 possible roots, you can a) try them all algorithmically, or b) apply some heuristics. Graph the polynomial (preferably on a calculator) and look at which of the roots seem to be true (where the funtion intersects the X-Axis)--After that, try each of your roots until you find the root you're looking for.

    Divide the equation by your new-found root and find the other two from your quadratic.

    Hope I helped.
     
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