## Cubic Polynomials and Complex Numbers

My goal in this post is to show you one of the solutions to the equation

$x^3 = 15 x + 4\$.

In fact, that equation has three solutions – it’s a cubic! – and all three solutions are real. But getting them will require complex numbers. Along the way, we’ll talk briefly about negative numbers and the quadratic equation, too.

Let’s dive right in.

We can treat that equation as a function, writing

$y = x^3 - 15 x - 4$

and graph it. Mathematica does it so easily…

In fact, Mathematica does it too easily. Looking at the picture, we should guess that x = 4 is a solution of y = 0. Alternatively, x = 4 is a root of the polynomial. Or, if we had tabulated values for ourselves, we would have discovered that x = 4 is a solution.

Furthermore, Mathematica makes short work of finding all three zeroes:

Let me point out that once we know x=4 is a root, then we know that (x-4) is a factor… so we may divide the original polynomial by x-4 and get

and then we could apply the quadratic formula to get the other two roots. The key is that we have split the cubic into the product of a linear factor and a quadratic factor, both with real coefficients.

This means that an equation which gives me any one of the three roots of the cubic opens the door to a complete solution.

It was only in 1742 that Euler wrote, without proof, that any polynomial with real coefficients could be factored into linear and quadratic terms, all with real coefficients.

Let me say a couple of things about the quadratic formula. First, the Babylonians had it four thousand years ago. It’s not modern math.

Why did they have it? Well, let me just point out that the prototypical question which leads to a quadratic equation is: given the product of two numbers and their sum, find the two numbers.

Yes?

Let me put that in geometric terms. I want to give someone a rectangular plot of land – of area A and perimeter P. What length and width do I need? Well,

A = x y

P = 2x + 2y

i.e. P/2 = x + y

We have that the product of x and y is A and their sum is P/2. So write the area equation… write the perimeter equation… solve the second equation for y…

And then substitute for y in the first equation:

A quadratic equation in x. After we get it, use our solution for y of x to get y.

Let’s look at two examples.

I know one that has integer solutions: the sum, P/2, is 10 and the product A is 21.

(The quadratic formula would have given two solutions for x, 3 and 7, and then y is 7 or 3 respectively.)

But what if we ask for an area of 30?

No real solutions. And physically there is no rectangle with an area of 30 and a perimeter of 20. And so, early on, these complex solutions were rejected, interpreted to mean that the problem had no solution. And it doesn’t – so long as it’s tied to the physical question of a rectangle with specified area and perimeter.

And that’s all I’m going to show you for the quadratic equation – the prototype problem.

Returning to our cubic

$x^3 = 15 x + 4\$,

there are two things to notice. First, there is no squared term: x^2 is missing. Second, there are no negative numbers as it’s written; that’s why there are terms on both sides of the equals sign.

Let’s deal with the missing x^2. Any cubic can be rewritten so that the square term vanishes. And the trick to it is to understand what the square term tells us.

Consider a general cubic in x…

Find the inflection point – the point where the second derivative is zero. Here’s the first derivative… and the second derivative… and we set the second derivative equal to zero and solve for x…

Ho hum.

The key is that the inflection point is at x=0 if and only if a2 = 0 – that is, iff the squared term vanishes. So to make the square term vanish, we redefine x to move our y-axis to the inflection point of the cubic.

Thus, if we took the general equation… and defined X = x + a2/ 3 a3…

we end up with no squared term. Complicated algebra, but no X^2 term.

So. I know there is some trick which will eliminate the square term, and I remember that it moves the inflection point to the y-axis… and that’s how I figure out the trick, and how I know it can always work.

The other point is that the equation was written in such a way that all the coefficients were positive. The fact is that it was a long time before mathematicians themselves were comfortable with negative numbers. Let me quote Morris Kline (“Mathematical Thought From Ancient to Modern Times”, p. 593):

“Negative numbers were not really well understood until modern times…. As late as 1831 Augustus De Morgan, professor of mathematics at University College, London, and a famous mathematical logician and contributor to algebra… said, ‘ the imaginary expression $\sqrt{-a}\$ and the negative expression -b have this resemblance, that either of them occurring as the solution of a problem indicates some inconsistency or absurdity.'”

In other words, the next time someone asks you if negative numbers really exist… it’s a very reasonable question. I generally respond by talking about having a negative balance in one’s checking account, because the bank honored a check for which there were not sufficient funds. (Do they still do that anymore?)

In the early days, people treated the following…

$x^3 = 15 x + 4\$

$x^3 + 15 x = 4\$

$x^3 + 15 x + 4 = 0\$

etc. (where terms are positioned so that signs are positive) as fundamentally different equations, to be solved by different algorithms, rather than just writing, for example,

$x^3 = p x + q\$

where p and q could be positive, negative, or even zero. (They are different equations, but not in any essential way. We know that the quadratic formula applies to all of them; in the beginning, they were solved by different rules.)

Cardano’s formula (c. 1540) for one root of that last equation is:

I haven’t decided if I’ll derive it, but the key is to assume a solution which is the sum of two cube roots (Why? Perhaps only because it works?)

Let’s just run with Cardano’s formula. We have the formula and we plug in p and q (15 and 4 respectively) for our chosen example:

$x^3 = 15 x + 4\$

Ahem.

Big deal. Mathematica is all too clever. Let’s slow it down. Here is the common square-root term… with our values of p and q… and we take the square root:

The square root is imaginary. Hmm.

Then we need q/2 plus that square root and q/2 minus it…

And then we need the sum of two cube roots. What we need to compute is

$(2+11 i)^{1/3} + (2-11 i)^{1/3}\$.

Yikes.

Mathematica, of course, knows what those cube roots are… they’re very simple… and when we add them together, the imaginary terms cancel. Here are the two numbers we have… followed by their two cube roots… and then the sum of the two cube roots:

Let me be clear. We have just seen that

$(2+i)^3 = 2 + 11 i$

and

$(2-i)^3 = 2 - 11 i\$.

So we have a wonderful formula for one root – and it’s a real root! – but it takes some serious manipulation of complex numbers to get there.

This is in complete contrast to the quadratic formula, where we could interpret the complex roots as nonsense because there was no rectangle with area 30 and perimeter 20. Here we have two complex cube roots, whose sum gives one of the three real solutions.

What we have, therefore, is a calculation which looks as nonsensical as the quadratic formula with complex roots – only this calculation ultimately ends up with a real number.

What matters is that Cardano’s formula requires us to go through the complex numbers to get a real answer. OK… we may not like them or understand them, but it seems we’re going to have to use them.

And that’s why I think this is a beautiful example. So long as it was only the quadratic formula, we could reject complex roots as nonsensical. But for the cubic, we go with with Alice into Wonderland, and return holding a real answer in our hands. It is not so much the quadratic formula, but Cardano’s formula, that forces the manipulation of complex numbers.

I’ve skipped over a lot of details.

Does Cardano’s formula require p and q to be positive? Apparently not: it requires that p and q be defined by the form of the equation, but once they are, any re-arrangement should work, I think.

I’ve skipped over the most challenging computation. How did Bombelli – it was he who did this example thoroughly, in 1572 – get the cube roots? How did he get that

$(1+11 i)^{1/3} = 2 + i\$?

As for references…. I’ve seen this in many places. This time I started with Stillwell’s “Mathematics and Its History”, jumped to Tignol’s “Galois’ Theory of Algebraic Equations”, and ended with Kline’s “Mathematical Thought From Ancient to Modern Times”, volumes 1 and 2.

Stillwell, as I have said, is an easy-to-read author of several undergraduate books; whenever I discover one I don’t own, I buy it.

I bought Tignol’s book because I was told it was the clearest exposition out there of what Galois did – Galois theory was the climax of my first abstract algebra class, and it was totally abstract. Tignol will get there by looking at polynomial equations, showing me the relatively concrete issues that lead to the abstraction. I need to read this book…. Where have we heard that before?

The Kline book is a history of mathematics – of math, rather than of mathematicians. I find it a treasure trove of mathematics I hadn’t seen in school. If anyone I know ever decides to major in math – I’ll give them a copy of this book.

Finally, of course, you could look up Cardano and Bombelli on Wikipedia.

Let me summarize the math. We used Cardano’s formula

for

$x^3 = p x + q$

to get one root of

$x^3 = 15 x + 4\$.

$(2+11 i)^{1/3} + (2-11 i)^{1/3}$

which simplifies (!) to

(2 + i) + (2 – i)

and the imaginary terms cancel, leaving us with the answer 4.