The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= \frac{ 1 }{ 2 }\\[1 em]x_3 &= -1 \end{aligned} $$Step 1:
Factor out $ \color{blue}{ x }$ from $ 2x^3+x^2-x $ and solve two separate equations:
$$ \begin{aligned} 2x^3+x^2-x & = 0\\[1 em] \color{blue}{ x }\cdot ( 2x^2+x-1 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 2x^2+x-1 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ 2x^2+x-1 = 0 $ are: $ x = -1 ~ \text{and} ~ x = \dfrac{ 1 }{ 2 }$.
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