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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


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    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


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    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
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