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

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

Factor the left side of the equation.

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

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

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

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

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

Factor $a$ out of $4 a$ .

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

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

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

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

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

Factor using the perfect square rule.

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

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

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

$4 a = 2 \cdot a \cdot 2$

Rewrite the polynomial.

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

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

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

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

$a = 0$

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

Set $a$ equal to $0$ .

$a = 0$

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

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

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

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

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

$a - 2 = 0$

Add $2$ to both sides of the equation.

$a = 2$

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

$a = 0, 2$