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

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 1.7656\\[1 em]x_3 &= -2.2656 \end{aligned} $$

Step 1:

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

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

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

Step 2:

The solutions of $ 2x^2+x-8 = 0 $ are: $ x = -\dfrac{ 1 }{ 4 }-\dfrac{\sqrt{ 65 }}{ 4 } ~ \text{and} ~ x = -\dfrac{ 1 }{ 4 }+\dfrac{\sqrt{ 65 }}{ 4 }$.

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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