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

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= \frac{ 3 }{ 2 }\\[1 em]x_3 &= -1 \end{aligned} $$

Step 1:

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

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

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

Step 2:

The solutions of $ 2x^2-x-3 = 0 $ are: $ x = -1 ~ \text{and} ~ x = \dfrac{ 3 }{ 2 }$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

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