Solve the Inequality for x (x+1)/((x-4)(x+8))>=0

$\frac{x + 1}{(x - 4) (x + 8)} \geq 0$

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

$x + 1 = 0$

$x - 4 = 0$

$x + 8 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Add $4$ to both sides of the equation.

$x = 4$

Subtract $8$ from both sides of the equation.

$x = - 8$

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

$x = - 1$

$x = 4$

$x = - 8$

Consolidate the solutions.

$x = - 1, 4, - 8$

Find the domain of $\frac{x + 1}{(x - 4) (x + 8)}$ .

Set the denominator in $\frac{x + 1}{(x - 4) (x + 8)}$ equal to $0$ to find where the expression is undefined.

$(x - 4) (x + 8) = 0$

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 $0$ .

$x - 4 = 0$

$x + 8 = 0$

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.

$x = 4$

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 $(x - 4) (x + 8) = 0$ true.

$x = 4, - 8$

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

$(- \infty, - 8) \cup (- 8, 4) \cup (4, \infty)$

Use each root to create test intervals.

$x < - 8$

$- 8 < x < - 1$

$- 1 < x < 4$

$x > 4$

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 $x < - 8$ to see if it makes the inequality true.

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$\frac{(- 10) + 1}{((- 10) - 4) ((- 10) + 8)} \geq 0$

The left side $- 0.32142857$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 8 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 8 < x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

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

$\frac{(- 4) + 1}{((- 4) - 4) ((- 4) + 8)} \geq 0$

The left side $0.09375$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{(0) + 1}{((0) - 4) ((0) + 8)} \geq 0$

The left side $- 0.03125$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

$\frac{(6) + 1}{((6) - 4) ((6) + 8)} \geq 0$

The left side $0.25$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 8$ False

$- 8 < x < - 1$ True

$- 1 < x < 4$ False

$x > 4$ True

$x < - 8$ False

$- 8 < x < - 1$ True

$- 1 < x < 4$ False

$x > 4$ True

The solution consists of all of the true intervals.

$- 8 < x \leq - 1$ or $x > 4$

The result can be shown in multiple forms.

Inequality Form:

$- 8 < x \leq - 1 or x > 4$

Interval Notation:

$(- 8, - 1] \cup (4, \infty)$

Solve the Inequality for x (x+1)/((x-4)(x+8))>=0

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