Solve by Factoring x^3-4x^2=0

$x^{3} - 4 x^{2} = 0$

Step 1

Factor $x^{2}$ out of $x^{3} - 4 x^{2}$ .

Factor $x^{2}$ out of $x^{3}$ .

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

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

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

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

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

Step 2

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

$x^{2} = 0$

$x - 4 = 0$

Step 3

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

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

$x^{2} = 0$

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

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Step 4

Set $x - 4$ equal to $0$ and solve for $x$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 5

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

$x = 0, 4$