Find the Roots (Zeros) f(x)=x^3-8x^2+16x

$f (x) = x^{3} - 8 x^{2} + 16 x$

Step 1

Set $x^{3} - 8 x^{2} + 16 x$ equal to $0$ .

$x^{3} - 8 x^{2} + 16 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 8 x^{2} + 16 x$ .

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

$x \cdot x^{2} - 8 x^{2} + 16 x = 0$

Factor $x$ out of $- 8 x^{2}$ .

$x \cdot x^{2} + x (- 8 x) + 16 x = 0$

Factor $x$ out of $16 x$ .

$x \cdot x^{2} + x (- 8 x) + x \cdot 16 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 8 x)$ .

$x (x^{2} - 8 x) + x \cdot 16 = 0$

Factor $x$ out of $x (x^{2} - 8 x) + x \cdot 16$ .

$x (x^{2} - 8 x + 16) = 0$

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$x (x^{2} - 8 x + 4^{2}) = 0$

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

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

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

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

$x = 0$

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

Set $x$ equal to $0$ .

$x = 0$

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 4)}^{2}$ equal to $0$ .

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

Solve ${(x - 4)}^{2} = 0$ for $x$ .

Set the $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $x {(x - 4)}^{2} = 0$ true.

$x = 0, 4$

Step 3