Solve the Inequality for x x^3+8x^2<0

$x^{3} + 8 x^{2} < 0$

Convert the inequality to an equation.

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

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

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

$x^{2} x + 8 x^{2} = 0$

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

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

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

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

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

$x^{2} = 0$

$x + 8 = 0$

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 square 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 $0$ .

$x = 0$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $x^{2} (x + 8) = 0$ true.

$x = 0, - 8$

The solution consists of all of the true intervals.

$x < - 8$

The result can be shown in multiple forms.

Inequality Form:

$x < - 8$

Interval Notation:

$(- \infty, - 8)$