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

$f (x) = x^{3} - 2 x^{2} + x$

Step 1

Set $x^{3} - 2 x^{2} + x$ equal to $0$ .

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

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

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

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

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

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

Raise $x$ to the power of $1$ .

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

Factor $x$ out of $x^{1}$ .

$x \cdot x^{2} + x (- 2 x) + x \cdot 1 = 0$

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

$x (x^{2} - 2 x) + x \cdot 1 = 0$

Factor $x$ out of $x (x^{2} - 2 x) + x \cdot 1$ .

$x (x^{2} - 2 x + 1) = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$x (x^{2} - 2 x + 1^{2}) = 0$

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

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x (x^{2} - 2 \cdot x \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

$x {(x - 1)}^{2} = 0$

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

$x = 0$

${(x - 1)}^{2} = 0$

Set $x$ equal to $0$ .

$x = 0$

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

${(x - 1)}^{2} = 0$

Solve ${(x - 1)}^{2} = 0$ for $x$ .

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $x {(x - 1)}^{2} = 0$ true.

$x = 0, 1$

Step 3