<s>[INST] Finding the roots of $z^4+z^3+z^2+z+1$ I need to find the roots of $z^4+z^3+z^2+z+1$. One approach I have is to guess the first linear solution, then use polynomial long division to find a 3rd degree polynomial, guess another from that, long division again, then use the quadratic formula to solve the quadratic. 
I think this will work but seems a bit work-heavy. Is there a simpler way?
 [/INST] In addition to the multiply-by-$\left(z - 1\right)$ trick, there is also the "symmetric" method to solve for the roots directly:
$$z^4 + z^3 + z^2 + z + 1 = z^2 \left(z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2}\right) = 0 $$
Since $z^2 \ne 0$, we can cancel it out, and we recognize the almost-square $z^2 + \frac{1}{z^2}$:
$$z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2} = \left(z + \frac{1}{z}\right)^2 - 2 + \left(z + \frac{1}{z}\right) + 1 = 0 $$
Now we can substitute $y = z + \frac{1}{z}$:
$$y^2 + y - 1 = 0 \to y = \frac{1\pm\sqrt{5}}{2}$$
We also have:
$$z + \frac{1}{z} = y \to z^2 + 1 = yz \to z=\frac{y\pm\sqrt{y^2 - 4}}{2}$$
We plug in $ y = \frac{1\pm\sqrt{5}}{2} $ into the equation above, and the reader verifies that the answers are equivalent to:
$$e^{i\frac{2\pi k}{5}}=\cos{\frac{2\pi k}{5}} + i\sin{\frac{2\pi k}{5}}, 0\le k\le 4$$
as obtained from the multiply-by-$\left(z-1\right)$ method.
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