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}{ 12x }$ from $ 12x^3-132x^2+336x $ and solve two separate equations:
$$ \begin{aligned} 12x^3-132x^2+336x & = 0\\[1 em] \color{blue}{ 12x }\cdot ( x^2-11x+28 ) & = 0 \\[1 em] \color{blue}{ 12x = 0} ~~ \text{or} ~~ x^2-11x+28 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ x^2-11x+28 = 0 $ are: $ x = 4 ~ \text{and} ~ x = 7$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.