Area of polygon on complex plane formed by complex roots of a polynomial 
Compute the area of the polygon whose vertices are the solutions in the complex plane to the polynomial $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$.

What I did was find the complex numbers by factoring and then using Surveyor's formula. However, that required me to use synthetic or polynomial division. I was wondering if there was a way to find the numbers without directly using division or whether there's a geometric approach to this?

Simply note that the polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
The is a unit square inside an octagon. Removing the square reveals $4$ little triangles. Thus the area of one little triangle is $\frac{2\sqrt{2} - 1}{4} = \frac{\sqrt{2}}{2} - \frac{1}{4}$.
$$2\sqrt{2} - \left(\frac{\sqrt{2}}{2} - \frac{1}{4}\right) = \frac{3\sqrt{2}}{2} + \frac{1}{4}$$
