Find roots of polynomial $$ p(x) = x^{20}+1+x^2 $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 0.6875+0.7482i\\[1 em]x_2 &= 0.6875-0.7482i\\[1 em]x_3 &= -0.3984+0.9049i\\[1 em]x_4 &= -0.3984-0.9049i\\[1 em]x_5 &= -0.1042+0.9265i\\[1 em]x_6 &= -0.1042-0.9265i\\[1 em]x_7 &= 0.9056+0.4911i\\[1 em]x_8 &= 0.9056-0.4911i\\[1 em]x_9 &= 0.1042+0.9265i\\[1 em]x_{10} &= 0.1042-0.9265i\\[1 em]x_{11} &= -0.6875+0.7482i\\[1 em]x_{12} &= -0.6875-0.7482i\\[1 em]x_{13} &= 0.3984+0.9049i\\[1 em]x_{14} &= 0.3984-0.9049i\\[1 em]x_{15} &= 1.0223+0.1709i\\[1 em]x_{16} &= 1.0223-0.1709i\\[1 em]x_{17} &= -0.9056+0.4911i\\[1 em]x_{18} &= -0.9056-0.4911i\\[1 em]x_{19} &= -1.0223+0.1709i\\[1 em]x_{20} &= -1.0223-0.1709i \end{aligned} $$

Step 1:

Write polynomial in descending order

$$ \begin{aligned} x^{20}+1+x^2 & = 0\\[1 em] x^{20}+x^2+1 & = 0 \end{aligned} $$

Step 2:

Polynomial $ x^{20}+x^2+1 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.

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