# Descartes’ Method for Constructing Roots of Polynomials with ‘Simple’ Curves - 'Depressed' Quartics and Cubics

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

### How Descartes solved ‘depressed’ quartics and cubics by intersecting a parabola with a circle

By Descartes’ time, it was known that if you had a quartic function of the form $f(x)=x^{4}+bx^{3}+cx^{2}+dx+e,$ then $f(x-\frac{b}{4})$ would become a related ‘depressed’ quartic of the form $g(x)=x^{4}+px^{2}+qx+r.$ So Descartes described a process for constructing the roots of any depressed quartic by intersecting the parabola $y=x^{2}$ with a certain circle whose center and radius were functions of $p, q,$ and $r.$

The process works as follows (Boyer 97):

1. Beginning with the depressed quartic equation $x^{4}+px^{2}+qx+r=0,$ separate the term $px^2$ into $(p-1)x^2$ and $1x^2$ to get $x^{4}+(p-1)x^{2}+x^{2}+qx=-r.$
2. Substitute $y=x^2$ in the $x^4$ term and the first $x^2$ term to rewrite the equation as $y^{2}+(p-1)y+x^{2}+qx=-r.$
3. This is the equation of a circle. Now complete the square on the quadratics in $y$ and $x$ to write the equation as ${\left({y+\frac{p-1}{2}}\right)}^{2}+{\left({x+\frac{q}{2}}\right)}^{2} = -r+{\left({\frac{p-1}{2}}\right)}^{2}+{\left({\frac{q}{2}}\right)}^{2}.$
4. Intersecting this circle with the parabola $y=x^2$ will provide the (real) solutions to the original depressed cubic.

For example, the equation $x^{4}+4x^{3}-9x^{2}-16x+20=0$ can be depressed by replacing $x$ with $(x-1)$ to get an equation whose roots are each one greater than the roots of the original equation; namely, $(x-1)^{4}+4(x-1)^{3}-9(x-1)^{2}-16(x-1)+20=0,$ which simplifies to $x^{4}-15x^{2}+10x+24=0.$

1. Rewrite $x^{4}-15x^{2}+10x+24=0$ as $x^{4}-16x^{2}+x^{2}+10x=-24.$
2. Substitute $y=x^{2}$ in the first two terms to obtain $y^{2}-16y+x^{2}+10x=-24.$
3. Complete the square in $y$ and $x$ to obtain $(y-8)^{2}+(x+5)^{2}=-24+64+25=65.$
4. In Figure 5, below, are the graphs of the circle $(x+5)^{2}+(y-8)^{2}=65$ and the parabola $y=x^{2}.$ Notice that the $x$-coordinates of the four intersection points agree with the four roots $-4, -1, 2,$ and $3.$ Thus the four roots of the original equation were $-5, -2, 1,$ and $2.$
Figure 5. Constructing solutions to quartic equations. Instructions: Change the sliders for $p, q,$ and $r$ to see the real roots of the depressed quartic equation $x^4+px^2+qx+r=0.$ In the case of double roots the circle will be tangent to the parabola at one point. For imaginary roots, the circle and the parabola will not intersect at all.
To solve a depressed cubic, Descartes simply multiplied the equation through by $x$ and solved the resulting depressed quartic (ignoring the solution $x=0$).