Algebra Examples

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

Step 1

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

$x + 1 = 0$

$x - 4 = 0$

$x + 8 = 0$

Step 2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3

Add $4$ to both sides of the equation.

$x = 4$

Step 4

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 5

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

$x = - 1$

$x = 4$

$x = - 8$

Step 6

Consolidate the solutions.

$x = - 1, 4, - 8$

Step 7

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

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Step 7.1

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

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

Step 7.2

Solve for $x$ .

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Step 7.2.1

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

$x - 4 = 0$

$x + 8 = 0$

Step 7.2.2

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

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Step 7.2.2.1

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

$x - 4 = 0$

Step 7.2.2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 7.2.3

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

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Step 7.2.3.1

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

$x + 8 = 0$

Step 7.2.3.2

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 7.2.4

The final solution is all the values that make $(x - 4) (x + 8) = 0$ true.

$x = 4, - 8$

Step 7.3

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

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

Step 8

Use each root to create test intervals.

$x < - 8$

$- 8 < x < - 1$

$- 1 < x < 4$

$x > 4$

Step 9

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

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Step 9.1

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

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Step 9.1.1

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

$x = - 10$

Step 9.1.2

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

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

Step 9.1.3

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

False

Step 9.2

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

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Step 9.2.1

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

$x = - 4$

Step 9.2.2

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

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

Step 9.2.3

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

True

Step 9.3

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

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Step 9.3.1

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

$x = 0$

Step 9.3.2

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

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

Step 9.3.3

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

False

Step 9.4

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

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Step 9.4.1

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

$x = 6$

Step 9.4.2

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

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

Step 9.4.3

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

True

Step 9.5

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

Step 10

The solution consists of all of the true intervals.

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

Step 11

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 12