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

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 0.7665+0.6083i\\[1 em]x_3 &= 0.7665-0.6083i\\[1 em]x_4 &= -0.7665+0.8972i\\[1 em]x_5 &= -0.7665-0.8972i \end{aligned} $$

Step 1:

Write polynomial in descending order

$$ \begin{aligned} 2x^3-4x^2-3x^6 & = 0\\[1 em] -3x^6+2x^3-4x^2 & = 0 \end{aligned} $$

Step 2:

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

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

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

Step 3:

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

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