Find roots of polynomial $$ p(x) = x^{11}+x^{10}+2x^9+3x^8+x^7+x^6+8x^5+2x^4+x^3-x^2-8x $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 0.8493\\[1 em]x_3 &= -1.0358\\[1 em]x_4 &= -0.122+0.9993i\\[1 em]x_5 &= -0.122-0.9993i\\[1 em]x_6 &= 0.9349+0.8938i\\[1 em]x_7 &= 0.9349-0.8938i\\[1 em]x_8 &= 0.0416+1.6267i\\[1 em]x_9 &= 0.0416-1.6267i\\[1 em]x_{10} &= -1.2612+0.6595i\\[1 em]x_{11} &= -1.2612-0.6595i \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x }$ from $ x^{11}+x^{10}+2x^9+3x^8+x^7+x^6+8x^5+2x^4+x^3-x^2-8x $ and solve two separate equations:

$$ \begin{aligned} x^{11}+x^{10}+2x^9+3x^8+x^7+x^6+8x^5+2x^4+x^3-x^2-8x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^{10}+x^9+2x^8+3x^7+x^6+x^5+8x^4+2x^3+x^2-x-8 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^{10}+x^9+2x^8+3x^7+x^6+x^5+8x^4+2x^3+x^2-x-8 & = 0 \end{aligned} $$

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

Step 2:

Polynomial $ x^{10}+x^9+2x^8+3x^7+x^6+x^5+8x^4+2x^3+x^2-x-8 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.

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