$$2p(-2p^2+p+4) = 0$$
Answer
$$ \begin{matrix}p_1 = 0 & p_2 = \dfrac{ 1 }{ 4 }-\dfrac{\sqrt{ 33 }}{ 4 } & p_3 = \dfrac{ 1 }{ 4 }+\dfrac{\sqrt{ 33 }}{ 4 } \end{matrix} $$
Explanation
$$ \begin{aligned} 2p(-2p^2+p+4) &= 0&& \text{simplify left side} \\[1 em]-4p^3+2p^2+8p &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -4x^{3}+2x^{2}+8x = 0 } $, first we need to factor our $ x $.
$$ -4x^{3}+2x^{2}+8x = x \left( -4x^{2}+2x+8 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -4x^{2}+2x+8 = 0$.
$ -4x^{2}+2x+8 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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