The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -\frac{ 112 }{ 3 } \end{aligned} $$Step 1:
Factor out $ \color{blue}{ x^2 }$ from $ 3x^3+112x^2 $ and solve two separate equations:
$$ \begin{aligned} 3x^3+112x^2 & = 0\\[1 em] \color{blue}{ x^2 }\cdot ( 3x+112 ) & = 0 \\[1 em] \color{blue}{ x^2 = 0} ~~ \text{or} ~~ 3x+112 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
To find the last zero, solve equation $ 3x+112 = 0 $
$$ \begin{aligned} 3x+112 & = 0 \\[1 em] 3 \cdot x & = -112 \\[1 em] x & = - \frac{ 112 }{ 3 } \end{aligned} $$