The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 7\\[1 em]x_3 &= -4 \end{aligned} $$Step 1:
Factor out $ \color{blue}{ 2x^3 }$ from $ 2x^5-6x^4-56x^3 $ and solve two separate equations:
$$ \begin{aligned} 2x^5-6x^4-56x^3 & = 0\\[1 em] \color{blue}{ 2x^3 }\cdot ( x^2-3x-28 ) & = 0 \\[1 em] \color{blue}{ 2x^3 = 0} ~~ \text{or} ~~ x^2-3x-28 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ x^2-3x-28 = 0 $ are: $ x = -4 ~ \text{and} ~ x = 7$.
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