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Math Help - Cubic function solving

  1. #1
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    Question Cubic function solving

    Hello
    Was just wondering if anyone could show me a SIMPLE way of solving a cubic function. I have looked up Cardano's method and it goes way over my head. I need help, please if anyone can help it would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by desdemoniagirl View Post
    Hello
    Was just wondering if anyone could show me a SIMPLE way of solving a cubic function. I have looked up Cardano's method and it goes way over my head. I need help, please if anyone can help it would be greatly appreciated.
    Without going into the justification of Cardano's method you should just be able to take the Cardano solution and substitute in your coefficients.

    For example if you have

    x^3 + ax^2 + bx + c = 0

    Then let
    p = b - \frac{a^2}{3}
    q = c + \frac{2a^3 - 9ab}{27}
    u=\sqrt[3]{-\frac{q}{2} \pm \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}

    Then the roots are given by

    x = -\frac{p}{3u} + u - \frac{a}{3}

    See for example the Wikipedia page for the Cubic Equation.
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  3. #3
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    If there were a "SIMPLE" way, why would we need guys like Cardano to invent a method?

    Listen to "Knowledge". As a good general rule, if you cannot find Rational Roots, punt! Use numerical methods and produce Roots to any desired level of accuracy. Do you REALLY need exact solutions anyway?

    Note: Any good Math software provides another possibility. MathCad and Mathematica come to mind immediately.
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  4. #4
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    You can try looking for rational roots. If x= m/n is a rational root, then one factor of the polynomial is (nx- m). Since that will multiply the other roots, assuming your polynomial has integer coefficients, n must evenly divide the leading coefficient and m must evenly divide the constant term.

    That is, if m/n is a rational root of ax^3+ bx^2+ cx+ d= 0 then n must evenly divide a and m must evenly divide d. You can try constructing all m/n where m is a factor of d and n a factor of a (remembering to try both positive and negative numbers) to see if there are any rational roots. If there is one, dividing the polynomial by (x- m/n) or (nx-m) reduces the problem to a quadratic equation for the other roots.

    Of course, there may NOT be any rational solutions in which case either Cardano's method or a numerical solution must be used.
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  5. #5
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    Any particular reason my post suggesting the use of synthetic division was deleted?

    Synthetic Division employs the use of the Rational Root Theorem (p/q) to quickly determine roots (the real rational ones) and factor nth degree polynomials in a simplistic way suitable for pre-algebra and algebra students.
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