$$6x^6-5x^5+4x^4-3x^3+2x^2+x+0 = 0$$

$$ \begin{matrix}x_1 = 0 & x_2 = -0.29409 & x_3 = -0.17268+0.8324i \\[1 em] x_4 = -0.17268-0.8324i & x_5 = 0.73639+0.49182i & x_6 = 0.73639-0.49182i \end{matrix} $$

In order to solve $ \color{blue}{ 6x^{6}-5x^{5}+4x^{4}-3x^{3}+2x^{2}+x = 0 } $, first we need to factor our $ x $.

$$ 6x^{6}-5x^{5}+4x^{4}-3x^{3}+2x^{2}+x = x \left( 6x^{5}-5x^{4}+4x^{3}-3x^{2}+2x+1 \right) $$

$ x = 0 $ is a root of multiplicity $ 1 $.

The remaining roots can be found by solving equation $ 6x^{5}-5x^{4}+4x^{3}-3x^{2}+2x+1 = 0$.

This polynomial has no rational roots that can be found using Rational Root Test.

Roots were found using Newton method.

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