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The standard method of solving a quadratic equation (ax2+bx+c=0) is with the Quadratic Formula. A couple of people have asked me if there is a similar method for cubic equations (mz3+nz2+oz+p=0). Such an equation has three solutions, at least one of which is real. You probably should try to factor the equation first, as it may have at least one relatively easy solution. For example x3-2x2-2x-3=0 has these likely factors: x+1, x-1, x+3, and x-3. Dividing x3-2x2-2x-3 by each of them shows that x3-2x2-2x-3=(x2+x+1)(x-3). So 3 is a solution of x3-2x2-2x-3=0, and the other two can be found using the quadratic formula. Graphing may also give useful clues toward the solutions.
There are several general, exact methods for solving cubic equations, all of which are a little bit difficult. So a popular method is to actually estimate (as accurately as we want) the three solutions with one of several computer algorithms. Here is one of the exact methods:
Before we begin, we can transform mz3+nz2+oz+p=0 into the standard form x3+ax+b=0 (sometimes given as x3+ax=b with an opposite sign on b). Divide mz3+nz2+oz+p=0 by m to get the equation into the form z3+fz2+gz+h=0, then substitute x-f/3 for z to arrive at the form x3+ax+b=0 [where a=(3f-g2)/3 and b=(2f3-9fg+27h)/27]. Once we have solved for x, we can go back and deduce z.
Let A=cuberoot(-b/2+sqr(b2/4+a3/27))
and B=cuberoot(-b/2-sqr(b2/4+a3/27))Then the three solutions are:
x=A+B, x=-(A+B)/2+(A-B)sqr(-3)/2, and x=-(A+B)/2-(A-B)sqr(-3)/2