Find the Roots (Zeros) f(x)=x^3-12x^2+36x

$f (x) = x^{3} - 12 x^{2} + 36 x$

Step 1

Set $x^{3} - 12 x^{2} + 36 x$ equal to $0$ .

$x^{3} - 12 x^{2} + 36 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 12 x^{2} + 36 x$ .

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

$x \cdot x^{2} - 12 x^{2} + 36 x = 0$

Factor $x$ out of $- 12 x^{2}$ .

$x \cdot x^{2} + x (- 12 x) + 36 x = 0$

Factor $x$ out of $36 x$ .

$x \cdot x^{2} + x (- 12 x) + x \cdot 36 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 12 x)$ .

$x (x^{2} - 12 x) + x \cdot 36 = 0$

Factor $x$ out of $x (x^{2} - 12 x) + x \cdot 36$ .

$x (x^{2} - 12 x + 36) = 0$

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$x (x^{2} - 12 x + 6^{2}) = 0$

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

$12 x = 2 \cdot x \cdot 6$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 6$ .

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

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

$x = 0$

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

Set $x$ equal to $0$ .

$x = 0$

Set ${(x - 6)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 6)}^{2}$ equal to $0$ .

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

Solve ${(x - 6)}^{2} = 0$ for $x$ .

Set the $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $x {(x - 6)}^{2} = 0$ true.

$x = 0, 6$

Step 3