Find roots of polynomial $$ p(x) = x^3-4x^2+4x $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 2 \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x }$ from $ x^3-4x^2+4x $ and solve two separate equations:

$$ \begin{aligned} x^3-4x^2+4x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^2-4x+4 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^2-4x+4 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.

Step 2:

The solution of $ ~ x^2-4x+4 = 0 ~$ is $~ x = 2 ~$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

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