Algebra Examples

$\frac{(x - 12) (x + 5)}{{(x - 3)}^{3}} \leq 0$

Step 1

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

$x - 12 = 0$

$x + 5 = 0$

${(x - 3)}^{3} = 0$

Step 2

Add $12$ to both sides of the equation.

$x = 12$

Step 3

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 4

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 5

Add $3$ to both sides of the equation.

$x = 3$

Step 6

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

$x = 12$

$x = - 5$

$x = 3$

Step 7

Consolidate the solutions.

$x = 12, - 5, 3$

Step 8

Find the domain of $\frac{(x - 12) (x + 5)}{{(x - 3)}^{3}}$ .

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Step 8.1

Set the denominator in $\frac{(x - 12) (x + 5)}{{(x - 3)}^{3}}$ equal to $0$ to find where the expression is undefined.

${(x - 3)}^{3} = 0$

Step 8.2

Solve for $x$ .

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Step 8.2.1

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 8.2.2

Add $3$ to both sides of the equation.

$x = 3$

Step 8.3

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

$(- \infty, 3) \cup (3, \infty)$

Step 9

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 3$

$3 < x < 12$

$x > 12$

Step 10

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 10.1

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

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Step 10.1.1

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

$x = - 8$

Step 10.1.2

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

$\frac{((- 8) - 12) ((- 8) + 5)}{{((- 8) - 3)}^{3}} \leq 0$

Step 10.1.3

The left side $- 0.04507888$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $- 5 < x < 3$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 5 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 10.2.2

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

$\frac{((0) - 12) ((0) + 5)}{{((0) - 3)}^{3}} \leq 0$

Step 10.2.3

The left side $2.\overline{2}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $3 < x < 12$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $3 < x < 12$ and see if this value makes the original inequality true.

$x = 8$

Step 10.3.2

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

$\frac{((8) - 12) ((8) + 5)}{{((8) - 3)}^{3}} \leq 0$

Step 10.3.3

The left side $- 0.416$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 10.4

Test a value on the interval $x > 12$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $x > 12$ and see if this value makes the original inequality true.

$x = 14$

Step 10.4.2

Replace $x$ with $14$ in the original inequality.

$\frac{((14) - 12) ((14) + 5)}{{((14) - 3)}^{3}} \leq 0$

Step 10.4.3

The left side $0.02854996$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.5

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

$x < - 5$ True

$- 5 < x < 3$ False

$3 < x < 12$ True

$x > 12$ False

$x < - 5$ True

$- 5 < x < 3$ False

$3 < x < 12$ True

$x > 12$ False

Step 11

The solution consists of all of the true intervals.

$x \leq - 5$ or $3 < x \leq 12$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 5 or 3 < x \leq 12$

Interval Notation:

$(- \infty, - 5] \cup (3, 12]$

Step 13