Solve the Inequality for x ((x+2)(x-3))/((x-2)(x+5))<=0

$\frac{(x + 2) (x - 3)}{(x - 2) (x + 5)} \leq 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x + 2 = 0$

$x - 3 = 0$

$x - 2 = 0$

$x + 5 = 0$

Step 2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3

Add $3$ to both sides of the equation.

$x = 3$

Step 4

Add $2$ to both sides of the equation.

$x = 2$

Step 5

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = - 2$

$x = 3$

$x = 2$

$x = - 5$

Step 7

Consolidate the solutions.

$x = - 2, 3, 2, - 5$

Step 8

Find the domain of $\frac{(x + 2) (x - 3)}{(x - 2) (x + 5)}$ .

Set the denominator in $\frac{(x + 2) (x - 3)}{(x - 2) (x + 5)}$ equal to $0$ to find where the expression is undefined.

$(x - 2) (x + 5) = 0$

Solve for $x$ .

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

$x - 2 = 0$

$x + 5 = 0$

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.

$x = 2$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

$x = 2, - 5$

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 5) \cup (- 5, 2) \cup (2, \infty)$

Step 9

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 2$

$- 2 < x < 2$

$2 < x < 3$

$x > 3$

Step 10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$\frac{((- 8) + 2) ((- 8) - 3)}{((- 8) - 2) ((- 8) + 5)} \leq 0$

The left side $2.2$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$\frac{((- 4) + 2) ((- 4) - 3)}{((- 4) - 2) ((- 4) + 5)} \leq 0$

The left side $- 2.\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{((0) + 2) ((0) - 3)}{((0) - 2) ((0) + 5)} \leq 0$

The left side $0.6$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Replace $x$ with $2.5$ in the original inequality.

$\frac{((2.5) + 2) ((2.5) - 3)}{((2.5) - 2) ((2.5) + 5)} \leq 0$

The left side $- 0.6$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{((6) + 2) ((6) - 3)}{((6) - 2) ((6) + 5)} \leq 0$

The left side $0.\overline{54}$ is greater than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 5$ False

$- 5 < x < - 2$ True

$- 2 < x < 2$ False

$2 < x < 3$ True

$x > 3$ False

$x < - 5$ False

$- 5 < x < - 2$ True

$- 2 < x < 2$ False

$2 < x < 3$ True

$x > 3$ False

Step 11

The solution consists of all of the true intervals.

$- 5 < x \leq - 2$ or $2 < x \leq 3$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$- 5 < x \leq - 2 or 2 < x \leq 3$

Interval Notation:

$(- 5, - 2] \cup (2, 3]$

Step 13

Solve the Inequality for x ((x+2)(x-3))/((x-2)(x+5))<=0

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