The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 24\\[1 em]x_3 &= 6 \end{aligned} $$Step 1:
Factor out $ \color{blue}{ x }$ from $ x^3-30x^2+144x $ and solve two separate equations:
$$ \begin{aligned} x^3-30x^2+144x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^2-30x+144 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^2-30x+144 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ x^2-30x+144 = 0 $ are: $ x = 6 ~ \text{and} ~ x = 24$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.