# Problem of solving the cubic function

1. Apr 14, 2014

### Martin Zhao

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!

2. Apr 14, 2014

### micromass

3. Apr 14, 2014

### 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.

4. Apr 15, 2014

### micromass

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

5. Apr 16, 2014

### disregardthat

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.

Last edited: Apr 16, 2014