Algebra Examples

$\frac{x^{3} (x - 2)}{{(x + 3)}^{2}} < 0$

Step 1

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

$x^{3} = 0$

$x - 2 = 0$

${(x + 3)}^{2} = 0$

Step 2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Step 3

Simplify $\sqrt[3]{0}$ .

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Step 3.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 3.2

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Step 4

Add $2$ to both sides of the equation.

$x = 2$

Step 5

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

$x + 3 = 0$

Step 6

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

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

$x = 0$

$x = 2$

$x = - 3$

Step 8

Consolidate the solutions.

$x = 0, 2, - 3$

Step 9

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

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Step 9.1

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

${(x + 3)}^{2} = 0$

Step 9.2

Solve for $x$ .

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Step 9.2.1

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

$x + 3 = 0$

Step 9.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 9.3

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

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

Step 10

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 0$

$0 < x < 2$

$x > 2$

Step 11

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 11.1

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

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Step 11.1.1

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

$x = - 6$

Step 11.1.2

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

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

Step 11.1.3

The left side $192$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 11.2

Test a value on the interval $- 3 < x < 0$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $- 3 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 11.2.2

Replace $x$ with $- 2$ in the original inequality.

$\frac{{(- 2)}^{3} ((- 2) - 2)}{{((- 2) + 3)}^{2}} < 0$

Step 11.2.3

The left side $32$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $0 < x < 2$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $0 < x < 2$ and see if this value makes the original inequality true.

$x = 1$

Step 11.3.2

Replace $x$ with $1$ in the original inequality.

$\frac{{(1)}^{3} ((1) - 2)}{{((1) + 3)}^{2}} < 0$

Step 11.3.3

The left side $- 0.0625$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 11.4

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

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Step 11.4.1

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

$x = 4$

Step 11.4.2

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

$\frac{{(4)}^{3} ((4) - 2)}{{((4) + 3)}^{2}} < 0$

Step 11.4.3

The left side $2.61224489$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 11.5

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

$x < - 3$ False

$- 3 < x < 0$ False

$0 < x < 2$ True

$x > 2$ False

$x < - 3$ False

$- 3 < x < 0$ False

$0 < x < 2$ True

$x > 2$ False

Step 12

The solution consists of all of the true intervals.

$0 < x < 2$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$0 < x < 2$

Interval Notation:

$(0, 2)$

Step 14