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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- Feb 5th 2009, 06:06 PMdesdemoniagirlCubic 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. - Feb 5th 2009, 06:21 PMRincewind
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

$\displaystyle x^3 + ax^2 + bx + c = 0$

Then let

$\displaystyle p = b - \frac{a^2}{3}$

$\displaystyle q = c + \frac{2a^3 - 9ab}{27}$

$\displaystyle u=\sqrt[3]{-\frac{q}{2} \pm \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$

Then the roots are given by

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

See for example the Wikipedia page for the Cubic Equation. - Feb 5th 2009, 06:32 PMTKHunny
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. - Feb 7th 2009, 05:29 AMHallsofIvy
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 $\displaystyle ax^3+ bx^2+ cx+ d= 0$ 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. - Feb 7th 2009, 06:58 AMKnowledge
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.