# Descartes’ Method for Constructing Roots of Polynomials with ‘Simple’ Curves - Sixth Degree Equations with Six Real Positive Roots

Author(s):
Gary Rubinstein (Stuyvesant High School)

About his process for finding roots of sixth degree polynomials, Descartes wrote (Smith 224):

This rule never fails nor does it admit of any exceptions.

Yet his one example is a specific type with just four real roots that seems somewhat contrived. Just as a circle can intersect a parabola in four points, it would seem logical that it should only be able to intersect the right branch of Descartes' ‘parabola of the second order’ in four points as well. But what if the polynomial has six real roots?

For example, the equation $x^{6}-6x^{5}-15x^{4}+88x^{3}+64x^{2}-168x+36=0$ has two negative roots and four positive roots. Descartes would have us replace $x$ with $x-4$ to get the equation $x^{6}-30x^{5}+345x^{4}-1912x^{3}+5248x^{2}-6440x+2500=0$ and then create the circle and right branch of the Cartesian parabola, confident that they will intersect six times at the proper places. Though a standard parabola can only intersect a circle in at most four places, the Cartesian parabola can, indeed, intersect a circle at six points.

Figure 12. Sextic equation with six real roots. Change the values of the roots to see how the circle and Cartesian parabola vary.

On the second to last page of 'The Geometry,' Descartes admitted (Smith 239):

[I]n many of these problems it may happen that the circle cuts the parabola of the second class so obliquely that it is hard to determine the exact point of intersection. In such cases this construction is not of practical value. The difficulty could easily be overcome by forming other rules analogous to these, which might be done in a thousand different ways.

Rather than hint at a thousand ways to resolve this issue, there was a simple one that Descartes seems to have chosen not to reveal. He could have relaxed his restriction on the roots having to all be positive and also utilized the left branch of the Cartesian parabola, which he described and used in Book II (Smith 87) in his solution to another problem known as the ‘five line locus problem.’

In the applet above, slide the offset all the way to the left so that each of the roots is reduced by $4.$ The new circle will intersect the right branch of the Cartesian parabola in four places and the left branch of the Cartesian parabola in two places.

The original equation was $x^{6}-6x^{5}-15x^{4}+88x^{3}+64x^{2}-168x+36=0.$

Using Descartes’ formulas, we get values $b=3, n=2, a=1, h=1,$ and $k=6,$ and radius ${\rm{IP}}=5.$ Also, point $(h,k)$ is the center of the circle, and the equation of the Cartesian parabola is $y=\frac{x^{2}}{2}-\frac{3x}{2}-1+\frac{3}{x}.$ When the offset is set to –4 in Figure 12, above, the intersections of the circle and two-branched Cartesian parabola give as the six roots of the original equation, $x\approx -3.30968,$ $x\approx -1.85769,$ $x\approx 0.244316,$ $x=1,$ $x\approx 4.15473,$ and $x\approx 5.76832.$

Gary Rubinstein (Stuyvesant High School), "Descartes’ Method for Constructing Roots of Polynomials with ‘Simple’ Curves - Sixth Degree Equations with Six Real Positive Roots," Convergence (March 2016)