Sextics

It is known that there are not general equations for polynomials of degree 5 or higher (see proof here), which is why Descartes’ method to solving sextics (polynomials of degree 6) is so intriguing to me.

Descartes’ Rule of Signs

Early on in The Geometry, Descartes explained his Rule of Signs. We still use this rule and method today. It helps us get an idea how many of each type of roots a polynomial has. It gives a range how many of each type of root a polynomial has (that is positive real, negative real, and complex). We use a table to organize this:

Positive Real	Negative Real	Complex

To find the upper bound of how many positive real roots there are, we are going to look at the polynomial given to us in the form where is a nonzero constant term. Count the number of sign changes between each term. This number goes in the positive real column and is the maximum number of positive real roots there will be.

To find the upper bound of how many negative real roots there are, we are going to look at the polynomial Simplify, and count the number of sign changes between each term. This number goes in the negative real column and is the maximum number of negative real roots there will be.

We will use the Fundamental Theorem of Algebra to fill in the complex column of our table. The Fundamental Theorem of Algebra states that a polynomial of degree n will have n solutions. So if you have a polynomial of degree 6, and you wrote 2 in the positive column, and 2 in the negative column, you will write 2 in the complex column because there will be a total of 6 roots. You now have the maximum number of complex roots.

To fill in the rest of the table, (to write all possibilities of solutions you could get), we will go down 2 in both the positive real column and the negative real column (if it permits; we could not get a negative amount of zeros). After we do this, we will adjust the complex column accordingly (so that way we will still have 6 total roots). We need to remember to list all combinations of possibilities.

You can see examples of Descartes’ Rule of Signs here.
I have reminded you of Descartes’ Rule of Signs because the first step to solving the roots of sextic polynomials (polynomials of degree 6) is to convert the sextic polynomial into a polynomial with all positive roots. You do this by plugging in into in the sixth degree polynomial, where c is larger than the amount of all the polynomial’s real negative roots. Remember we can find the maximum number of a polynomial’s real negative roots by using Descartes’ rule of signs by counting the number of sign changes in the polynomial (see above). So we will chose a value for c that is larger than the amount of all the polynomial’s real negative roots and plug in into our polynomial. (Rubinstein, 6)

Now we have a sextic polynomial, in the form , where is a negative coefficient, is a positive coefficient, is a negative coefficient, is a positive coefficient, is a negative coefficient, and is a positive constant, and (which is satisfied when we substituted in for . (Rubinstein, 6)

Let . Graph the Cartesian parabola: , and the circle where , and . The intersections of the Cartesian parabola and the circle will be the real roots of the converted sextic polynomial. (Rubinstein, 6)

But what do the roots of the ‘converted’ sextic tell us about the original sextic polynomial?
This is just like the last step of Descartes’ method for solving for the roots of quartic polynomials. We plug in our root in for x in (x-c). This will give us the actual root of the polynomial. (Rubinstein, 6)

See it in action!

There are some awesome applets that show this in play on this site:

http://www.maa.org/press/periodicals/convergence/descartes-method-for-constructing-roots-of-polynomials-with-simple-curves-sextic-and-quintic

Why does it work?

I will show why the intersection of this circle and this Cartesian parabola make up the real roots of the converted sextic polynomial.

First, start with the generic equations of the Cartesian parabola and the circle: