Find the Roots (Zeros) x^3-4x^2+4x=0

$x^{3} - 4 x^{2} + 4 x = 0$

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 4 x^{2} + 4 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 4 x^{2} + 4 x = 0$

Factor $x$ out of $- 4 x^{2}$ .

$x \cdot x^{2} + x (- 4 x) + 4 x = 0$

Factor $x$ out of $4 x$ .

$x \cdot x^{2} + x (- 4 x) + x \cdot 4 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 4 x)$ .

$x (x^{2} - 4 x) + x \cdot 4 = 0$

Factor $x$ out of $x (x^{2} - 4 x) + x \cdot 4$ .

$x (x^{2} - 4 x + 4) = 0$

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

$x (x^{2} - 4 x + 2^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$x (x^{2} - 2 \cdot x \cdot 2 + 2^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

$x {(x - 2)}^{2} = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

${(x - 2)}^{2} = 0$

Set $x$ equal to $0$ .

$x = 0$

Set ${(x - 2)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 2)}^{2}$ equal to $0$ .

${(x - 2)}^{2} = 0$

Solve ${(x - 2)}^{2} = 0$ for $x$ .

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $x {(x - 2)}^{2} = 0$ true.

$x = 0, 2$