# Problem of solving the cubic function

by Martin Zhao
Tags: cubic, function, solving
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 P: 8 Guys, I may need your help. There is a question saying that how to solve the cubic function in general form, which means that y=ax^3+bx^2+cx+d. How do you guys solve for x? To be honest, I have no idea of this question. Probably, it uses the same way as the quartic function. Thanks!
 Mentor P: 18,293
 P: 6 There are numerous ways of solving cubic functions, but the most efficient way would be to plot it using some type of software--most easily maple, MATLAB, or Mathematica. Once you do that, locate one of the roots. Use the root as a factor, and divide the cubic by that factor to obtain a quadratic. Quadratics are easy to solve, thus you can easily find the remaining two roots.
Mentor
P: 18,293
Problem of solving the cubic function

 Quote by AMenendez There are numerous ways of solving cubic functions, but the most efficient way would be to plot it using some type of software--most easily maple, MATLAB, or Mathematica. Once you do that, locate one of the roots. Use the root as a factor, and divide the cubic by that factor to obtain a quadratic. Quadratics are easy to solve, thus you can easily find the remaining two roots.
Well, if you're going to be content with numerical answers, then there are many good techniques to approximate the roots to a very high degree: http://en.wikipedia.org/wiki/Newton's_method
 Sci Advisor P: 1,807 For the cubic equation $ax^3+bx^2+cx+d=0$ (in your case the constant term is $d-y$, not $d$), try substituting $x = z +\frac{\gamma}{z}$, and solve for $z$ by choosing the constant $\gamma$ correctly. If fairly certain that for a good choice of $\gamma$ (it will become apparant what $\gamma$ must be) you will end up with a quadratic function in $z^2$. This way you may arrive at the formula yourself, it's a neat exercise. You probably need to be careful verifying your solution afterwards, as $z +\frac{\gamma}{z}$ is not defined everywhere, and does not attain all values. To make calculations easier, you can assume $a = 1$ first, and make the necessary modification afterwards.

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