Using partial information to factor $x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$

Question

I wish to find exact expressions for all roots of $p(x)=x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ By observing that for the roots $x_0 \pm iy_0, x_0 \approx -0.15883609808599033632, y_0 \approx 0.27511219196092896700,$ we have that $x_0$ is the unique real root of $r(x) = x^3+12x^2+8x+1,$ I was able to prove that all roots of the original sextic can be expressed in radicals. The process is as follows:

1. Divide $p(x+iy)$ by $r(x)$ to get $\frac{1}{8}x^3 + \frac{3}{16}x^2 + x\left(\frac{7}{32}-\frac{15y^2}{8}\right) + \left(\frac{95}{32}-\frac{15y^2}{16}\right) + \frac{R(x,y)}{p(x)}$ where $R(x,y) = A(y)x^2 + B(y)x + C(y)$ and $A(y) = 15y^4 - \frac{15y^2}{4} - \frac{201}{16}, B(y) = 15y^8 - 30y^6 + 12y^4 + \frac{75y^2}{8} - \frac{767}{32}, C(y) = -y^6+5y^4-\frac{193y^2}{16}-\frac{63}{32}.$

2. The equation $R(x_0, y_0) = 0$ is a quartic in $y_0^2,$ which we can solve exactly to obtain $y_0^2$ and hence $y_0.$

3. Polynomial division reduces $p(x)$ to a quartic, and now we apply the quartic formula again to find the other $4$ roots.

However, I don't want to perform the rest of the computations. Is there a cleaner way to use the observation that $r(x_0) = 0,$ perhaps in the realm of abstract algebra?

Answer 1

The hint.

Use the following: $$x^6+3x^5+5x^4+10x^3+13x^2+4x+1=\left(x^3+\frac{3}{2}x^2-2x-1\right)^2+\frac{3}{4}x^2(3x+4)^2.$$ Now, you can get all roots of the polynomial. Just solve two cubic equations.

Answer 2

Just for the fun, starting from @Michael Rozenberg's answer, we need to solve the two cubic equations $$x^3+\frac{3}{2} \left(1-i \sqrt{3}\right) x^2-2 \left(1+i \sqrt{3}\right) x-1=0$$ $$x^3+\frac{3}{2} \left(1+i \sqrt{3}\right) x^2-2 \left(1-i \sqrt{3}\right) x-1=0$$

Prepare yourself for nasty solutions.

Answer 3

$$\frac{x^9+6x^6+31x^3-1}{x^3-3x^2+4x-1}$$