The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 1\\[1 em]x_3 &= \frac{ 1 }{ 2 } \end{aligned} $$Step 1:
Factor out $ \color{blue}{ 2x }$ from $ 4x^3-6x^2+2x $ and solve two separate equations:
$$ \begin{aligned} 4x^3-6x^2+2x & = 0\\[1 em] \color{blue}{ 2x }\cdot ( 2x^2-3x+1 ) & = 0 \\[1 em] \color{blue}{ 2x = 0} ~~ \text{or} ~~ 2x^2-3x+1 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ 2x^2-3x+1 = 0 $ are: $ x = \dfrac{ 1 }{ 2 } ~ \text{and} ~ x = 1$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.