Find roots of polynomial $$ p(x) = x^6+x^5+3x^4+3x^3+3x^2+2x $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -0.8163\\[1 em]x_3 &= -0.3433+1.0771i\\[1 em]x_4 &= -0.3433-1.0771i\\[1 em]x_5 &= 0.2514+1.3616i\\[1 em]x_6 &= 0.2514-1.3616i \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x }$ from $ x^6+x^5+3x^4+3x^3+3x^2+2x $ and solve two separate equations:

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

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

Step 2:

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

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