Find the Roots (Zeros) y=x^3-4x

$y = x^{3} - 4 x$

Set $x^{3} - 4 x$ equal to $0$ .

$x^{3} - 4 x = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 4 x$ .

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

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

Factor $x$ out of $- 4 x$ .

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

Factor $x$ out of $x \cdot x^{2} + x \cdot - 4$ .

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

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

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$x ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$x (x + 2) (x - 2) = 0$

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

$x = 0$

$x + 2 = 0$

$x - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $x (x + 2) (x - 2) = 0$ true.

$x = 0, - 2, 2$