Find the Roots (Zeros) 2x(x^2-36)=0

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

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

$x = 0$

$x^{2} - 36 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x^{2} - 36$ equal to $0$ and solve for $x$ .

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

$x^{2} - 36 = 0$

Solve $x^{2} - 36 = 0$ for $x$ .

Add $36$ to both sides of the equation.

$x^{2} = 36$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

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

$x = \pm \sqrt{6^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 6$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 6$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 6$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 6, - 6$

The final solution is all the values that make $2 x (x^{2} - 36) = 0$ true.

$x = 0, 6, - 6$