Find the Roots (Zeros) f(x)=x^3-6x^2+9x

$f (x) = x^{3} - 6 x^{2} + 9 x$

Step 1

Set $x^{3} - 6 x^{2} + 9 x$ equal to $0$ .

$x^{3} - 6 x^{2} + 9 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 6 x^{2} + 9 x$ .

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

$x \cdot x^{2} - 6 x^{2} + 9 x = 0$

Factor $x$ out of $- 6 x^{2}$ .

$x \cdot x^{2} + x (- 6 x) + 9 x = 0$

Factor $x$ out of $9 x$ .

$x \cdot x^{2} + x (- 6 x) + x \cdot 9 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 6 x)$ .

$x (x^{2} - 6 x) + x \cdot 9 = 0$

Factor $x$ out of $x (x^{2} - 6 x) + x \cdot 9$ .

$x (x^{2} - 6 x + 9) = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$x (x^{2} - 6 x + 3^{2}) = 0$

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

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$x (x^{2} - 2 \cdot x \cdot 3 + 3^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

$x {(x - 3)}^{2} = 0$

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

$x = 0$

${(x - 3)}^{2} = 0$

Set $x$ equal to $0$ .

$x = 0$

Set ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

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

${(x - 3)}^{2} = 0$

Solve ${(x - 3)}^{2} = 0$ for $x$ .

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $x {(x - 3)}^{2} = 0$ true.

$x = 0, 3$

Step 3