Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Factoring problem that has me stumped!

  1. May 26, 2004 #1

    [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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook