The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 5.8642\\[1 em]x_3 &= -1.3642 \end{aligned} $$Step 1:
Factor out $ \color{blue}{ 2x }$ from $ 4x^3-18x^2-32x $ and solve two separate equations:
$$ \begin{aligned} 4x^3-18x^2-32x & = 0\\[1 em] \color{blue}{ 2x }\cdot ( 2x^2-9x-16 ) & = 0 \\[1 em] \color{blue}{ 2x = 0} ~~ \text{or} ~~ 2x^2-9x-16 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ 2x^2-9x-16 = 0 $ are: $ x = \dfrac{ 9 }{ 4 }-\dfrac{\sqrt{ 209 }}{ 4 } ~ \text{and} ~ x = \dfrac{ 9 }{ 4 }+\dfrac{\sqrt{ 209 }}{ 4 }$.
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