Solve for x (x-5)^2(x+2)<0

${(x - 5)}^{2} (x + 2) < 0$

Step 1

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

${(x - 5)}^{2} = 0$

$x + 2 = 0$

Step 2

Set ${(x - 5)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 5)}^{2}$ equal to $0$ .

${(x - 5)}^{2} = 0$

Solve ${(x - 5)}^{2} = 0$ for $x$ .

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 3

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.

$x = - 2$

Step 4

The final solution is all the values that make ${(x - 5)}^{2} (x + 2) < 0$ true.

$x = 5, - 2$

Step 5

The solution consists of all of the true intervals.

$x < - 2$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x < - 2$

Interval Notation:

$(- \infty, - 2)$

Step 7