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

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 1.8081\\[1 em]x_3 &= -1.4748 \end{aligned} $$

Step 1:

Write polynomial in descending order

$$ \begin{aligned} 8x-3x^3+x^2 & = 0\\[1 em] -3x^3+x^2+8x & = 0 \end{aligned} $$

Step 2:

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

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

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

Step 3:

The solutions of $ 3x^2-x-8 = 0 $ are: $ x = \dfrac{ 1 }{ 6 }-\dfrac{\sqrt{ 97 }}{ 6 } ~ \text{and} ~ x = \dfrac{ 1 }{ 6 }+\dfrac{\sqrt{ 97 }}{ 6 }$.

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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