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

Having problem with roots

  1. Aug 10, 2012 #1
    Hello dear Physics Forums users.

    I m currently studying some Integrals from a book which my elder brother studied with years ago, and one of the problems had the denominator:


    Well, I m sure that's over my level, and teacher will never ask this, but in the solution it says its roots are x=1, x=2 and x=-3, without showing the steps, or saying how he found them.

    Wondering what the steps were, I typed the question at WolframAlpha, and it weirdly said step by step solution is unavailable.

    Can somebody show me a formula or a way to find the roots in such a question?
  2. jcsd
  3. Aug 10, 2012 #2
    In classroom settings, just plugging in obvious numbers often works. You can see right away that x = 1 works. 1 - 7 + 6 = 0. When confronted with a cubic or higher polynomial, always plug in x = 0, 1, -1 first thing.
  4. Aug 10, 2012 #3
    Hmm... I see. Will keep that in mind, thanks :)

    It would also be good to know some formal solution if there is one though :)
  5. Aug 10, 2012 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    There is no simple equivelent to the quadratic forumula for finding roots of a cubic equation. The sure fire method would be to graph the equation. This is now so easy, using a spreadsheet, that there is no real reason not to do it.

    Then look up something called the bisection method of root finding. Using that you can, starting from a rough guess on a graph, refine to any desired number of decimal places.
  6. Aug 10, 2012 #5
    I see...thanks for the answers guys :)
  7. Aug 10, 2012 #6
    I'll add here, just for completeness' sake, that there is a general equation to find the zeros of a cubic. However it's so nasty that it makes you understand why some people don't like math. I'll link to a page that has the formula, but I'll say I don't recommend actually using it. I'd do what Integral suggested, just saying this to expand knowledge.

  8. Aug 10, 2012 #7


    User Avatar
    Science Advisor

    First, whenever you look for a root, have the rational root test in your head. Specifically the only integer roots possible are the factors of the constant.

    The cubic ##x^3+px+q## has extra structure you can exploit:
    • if ##p + \frac{q}{k} = -k^2## then k is a root,
    • if ##p - \frac{q}{k} = -k^2## then -k is a root.
  9. Aug 10, 2012 #8
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook