Find roots of polynomial $$ p(x) = 3x^5+2x^4-x^3-8x^2+7x $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 0.84+0.3715i\\[1 em]x_3 &= 0.84-0.3715i\\[1 em]x_4 &= -1.1734+1.1785i\\[1 em]x_5 &= -1.1734-1.1785i \end{aligned} $$

Step 1:

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

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

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

Step 2:

Polynomial $ 3x^4+2x^3-x^2-8x+7 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.

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