Find roots of polynomial $$ p(x) = x^{10}+3x^9+3x^8-31x^7-48x^6+48x^5+122x^4+48x^3 $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -0.5605\\[1 em]x_3 &= -1.2394+0.471i\\[1 em]x_4 &= -1.2394-0.471i\\[1 em]x_5 &= 2.0126+0.1742i\\[1 em]x_6 &= 2.0126-0.1742i\\[1 em]x_7 &= -1.9929+2.822i\\[1 em]x_8 &= -1.9929-2.822i \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x^3 }$ from $ x^{10}+3x^9+3x^8-31x^7-48x^6+48x^5+122x^4+48x^3 $ and solve two separate equations:

$$ \begin{aligned} x^{10}+3x^9+3x^8-31x^7-48x^6+48x^5+122x^4+48x^3 & = 0\\[1 em] \color{blue}{ x^3 }\cdot ( x^7+3x^6+3x^5-31x^4-48x^3+48x^2+122x+48 ) & = 0 \\[1 em] \color{blue}{ x^3 = 0} ~~ \text{or} ~~ x^7+3x^6+3x^5-31x^4-48x^3+48x^2+122x+48 & = 0 \end{aligned} $$

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

Step 2:

Polynomial $ x^7+3x^6+3x^5-31x^4-48x^3+48x^2+122x+48 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.

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