Find roots of polynomial $$ p(x) = 12x^3-8x $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= \frac{\sqrt{ 6 }}{ 3 }\\[1 em]x_3 &= - \frac{\sqrt{ 6 }}{ 3 } \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ 4x }$ from $ 12x^3-8x $ and solve two separate equations:

$$ \begin{aligned} 12x^3-8x & = 0\\[1 em] \color{blue}{ 4x }\cdot ( 3x^2-2 ) & = 0 \\[1 em] \color{blue}{ 4x = 0} ~~ \text{or} ~~ 3x^2-2 & = 0 \end{aligned} $$

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

Step 2:

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

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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