Find roots of polynomial $$ p(x) = x^{20}+3x^{19}+x $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -3\\[1 em]x_3 &= 0.6056+0.7062i\\[1 em]x_4 &= 0.6056-0.7062i\\[1 em]x_5 &= -0.5986+0.7378i\\[1 em]x_6 &= -0.5986-0.7378i\\[1 em]x_7 &= -0.307+0.8924i\\[1 em]x_8 &= -0.307-0.8924i\\[1 em]x_9 &= 0.8068+0.4587i\\[1 em]x_{10} &= 0.8068-0.4587i\\[1 em]x_{11} &= 0.3318+0.8727i\\[1 em]x_{12} &= 0.3318-0.8727i\\[1 em]x_{13} &= -0.8223+0.4883i\\[1 em]x_{14} &= -0.8223-0.4883i\\[1 em]x_{15} &= 0.0157+0.938i\\[1 em]x_{16} &= 0.0157-0.938i\\[1 em]x_{17} &= 0.9132+0.1589i\\[1 em]x_{18} &= 0.9132-0.1589i\\[1 em]x_{19} &= -0.9452+0.1712i\\[1 em]x_{20} &= -0.9452-0.1712i \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x }$ from $ x^{20}+3x^{19}+x $ and solve two separate equations:

$$ \begin{aligned} x^{20}+3x^{19}+x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^{19}+3x^{18}+1 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^{19}+3x^{18}+1 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.

Step 2:

Polynomial $ x^{19}+3x^{18}+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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