Algebra Examples

$\frac{{(x - 5)}^{2} (x + 3)}{x - 6} \geq 0$

Step 1

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

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

$x + 3 = 0$

$x - 6 = 0$

Step 2

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

$x - 5 = 0$

Step 3

Add $5$ to both sides of the equation.

$x = 5$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Add $6$ to both sides of the equation.

$x = 6$

Step 6

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

$x = 5$

$x = - 3$

$x = 6$

Step 7

Consolidate the solutions.

$x = 5, - 3, 6$

Step 8

Find the domain of $\frac{{(x - 5)}^{2} (x + 3)}{x - 6}$ .

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Step 8.1

Set the denominator in $\frac{{(x - 5)}^{2} (x + 3)}{x - 6}$ equal to $0$ to find where the expression is undefined.

$x - 6 = 0$

Step 8.2

Add $6$ to both sides of the equation.

$x = 6$

Step 8.3

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

$(- \infty, 6) \cup (6, \infty)$

Step 9

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 5$

$5 < x < 6$

$x > 6$

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 $x < - 3$ to see if it makes the inequality true.

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Step 10.1.1

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

$x = - 6$

Step 10.1.2

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

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

Step 10.1.3

The left side $30.25$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 10.2

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

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Step 10.2.1

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

$x = 0$

Step 10.2.2

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

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

Step 10.2.3

The left side $- 12.5$ is less than the right side $0$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $5 < x < 6$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $5 < x < 6$ and see if this value makes the original inequality true.

$x = 5.5$

Step 10.3.2

Replace $x$ with $5.5$ in the original inequality.

$\frac{{((5.5) - 5)}^{2} ((5.5) + 3)}{(5.5) - 6} \geq 0$

Step 10.3.3

The left side $- 4.25$ is less than the right side $0$ , which means that the given statement is false.

False

Step 10.4

Test a value on the interval $x > 6$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 10.4.2

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

$\frac{{((8) - 5)}^{2} ((8) + 3)}{(8) - 6} \geq 0$

Step 10.4.3

The left side $49.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 10.5

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

$x < - 3$ True

$- 3 < x < 5$ False

$5 < x < 6$ False

$x > 6$ True

$x < - 3$ True

$- 3 < x < 5$ False

$5 < x < 6$ False

$x > 6$ True

Step 11

The solution consists of all of the true intervals.

$x \leq - 3$ or $x > 6$ or $x = 5$

Step 12

Combine the intervals.

$x \leq - 3 or x = 5 or x > 6$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or x = 5 or x > 6$

Interval Notation:

$(- \infty, - 3] \cup [5, 5] \cup (6, \infty)$

Step 14