We mathematician types like solving polynomial equations. The simplest such equations are the linear ones, meaning that the variable appears to the exponent one. They have the general form:

\[

ax+b=0.

\]

If you remember anything at all from your basic algebra classes, then you know that this is readily solved by bringing the *b* to the other side and dividing by *a*. We obtain

\[

x=\frac{-b}{a}.

\]

Of course, we are assuming here that *a* is not zero, but let’s not be overly pedantic. We can think of this as “the linear formula,” since it can be used to solve any linear equation we might confront. That seems a bit grandiose for something so simple, but what the heck.

As Stephen King once wrote, once you’ve done *Frankenstein*, what’s left to do but *Bride of Frankenstein*? And once you’ve done linear equations, you might as well move on to the quadratic case. That is, to equations where the exponent appears to the exponent two. They have the general form:

\[

ax^2+bx+c=0.

\]

Now, in your elementary algebra classes you learn that factoring is the preferred way of solving such things. But at some point we all must grow up and realize that factoring, so sorry, hardly ever works. So we move onto more sophisticated methods like “completing the square.” It goes like this. First we divide through by *a*, which ensures that the coefficient of the squared term will be one. That’s very convenient. Then we move the constant term to the other side to obtain:

\[

x^2+\frac{b}{a}(x)=-\frac{c}{a}.

\]

Now for the seemingly arbitrary step that has confused generations of middle school students. By adding the same thing to both sides we obtain this:

\[

x^2+\frac{b}{a}(x)+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}

\]

This was a clever thing to do, since that expression on the left-hand side is now a perfect square. Specifically, we have:

\[

\left( x+\frac{b}{2a} \right)^2 = -\frac{c}{a}+\frac{b^2}{4a^2}

\]

and

\[

x+\frac{b}{2a}=\pm \sqrt{ -\frac{c}{a}+\frac{b^2}{4a^2} }

\]

With a bit of elementary symbol manipulation this becomes:

\[

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a},

\]

which is the famous quadratic formula. It’s a very useful little gadget, since it shows you how to express the solutions of a quadratic equation as a simple function of its coefficients.

That was considerably more complicated than the linear case, but it really is nothing extraordinary. If you have some facility for mathematics the idea of completing the square is pretty natural. It’s the kind of thing a mathematically talented high school student might come up with. But when we kick things up to the cubic case, which looks like this:

\[

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

\]

things get considerably more complicated. There is no obvious way forward. There are, however, nonobvious ways forward, and, as we shall see, there is a cubic formula to go along with our linear and quadratic formulas. Since the full derivation has quite a few steps, we shall only do the first one this week. Notice, incidentally, that I have simply assumed that the coefficient of the cubic term is one. There is no loss of generality in making this assumption. If the leading coefficient is not one, then you can simply divide through by that coefficient to obtain an polynomial whose leading term is, indeed, one.

The first step is to make the following change of variables:

\[

x=y-\frac{a}{3}.

\]

That transforms our equation into this:

\[

\left(y-\frac{a}{3} \right)^3+a \left(y-\frac{a}{3} \right)^2+ b \left( y-\frac{a}{3} \right) + c =0

\]

It’s a bit tedious, but if you would care to multiply that out and group together the like terms, you will notice that a miracle happens. The squared term disappears! In other words, you get an equation that looks like this:

\[

y^3+py+q=0,

\]

where *p* and *q* are elaborate functions of the coefficients *a*, *b* and *c*. This sort of cubic equation, lacking a square term, is said to be a “reduced cubic.” If we could devise some general procedure for solving it, then we would be able to work our way backwards to solutions to the general cubic.

But does eliminating the square term really help us all that much? Indeed it does, as we shall explain in a future post.