Calculus Examples

$x^{3} + 2 x^{2} - 9 x - 18 > 0$

Step 1

Convert the inequality to an equation.

$x^{3} + 2 x^{2} - 9 x - 18 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor out the greatest common factor from each group.

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Step 2.1.1

Group the first two terms and the last two terms.

$(x^{3} + 2 x^{2}) - 9 x - 18 = 0$

Step 2.1.2

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x + 2) - 9 (x + 2) = 0$

Step 2.2

Factor the polynomial by factoring out the greatest common factor, $x + 2$ .

$(x + 2) (x^{2} - 9) = 0$

Step 2.3

Rewrite $9$ as $3^{2}$ .

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

Step 2.4

Factor.

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Step 2.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$(x + 2) ((x + 3) (x - 3)) = 0$

Step 2.4.2

Remove unnecessary parentheses.

$(x + 2) (x + 3) (x - 3) = 0$

Step 3

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

$x + 2 = 0$

$x + 3 = 0$

$x - 3 = 0$

Step 4

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 5.1

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

$x + 3 = 0$

Step 5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 6.1

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

$x - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 7

The final solution is all the values that make $(x + 2) (x + 3) (x - 3) = 0$ true.

$x = - 2, - 3, 3$

Step 8

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 2$

$- 2 < x < 3$

$x > 3$

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

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Step 9.1.1

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

$x = - 6$

Step 9.1.2

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

${(- 6)}^{3} + 2 {(- 6)}^{2} - 9 \cdot - 6 - 18 > 0$

Step 9.1.3

The left side $- 108$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $- 3 < x < - 2$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 3 < x < - 2$ and see if this value makes the original inequality true.

$x = - 2.5$

Step 9.2.2

Replace $x$ with $- 2.5$ in the original inequality.

${(- 2.5)}^{3} + 2 {(- 2.5)}^{2} - 9 \cdot - 2.5 - 18 > 0$

Step 9.2.3

The left side $1.375$ 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 $- 2 < x < 3$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $- 2 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 9.3.2

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

${(0)}^{3} + 2 {(0)}^{2} - 9 \cdot 0 - 18 > 0$

Step 9.3.3

The left side $- 18$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 9.4

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

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Step 9.4.1

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

$x = 6$

Step 9.4.2

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

${(6)}^{3} + 2 {(6)}^{2} - 9 \cdot 6 - 18 > 0$

Step 9.4.3

The left side $216$ 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.

$x < - 3$ False

$- 3 < x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

$x < - 3$ False

$- 3 < x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

Step 10

The solution consists of all of the true intervals.

$- 3 < x < - 2$ or $x > 3$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$- 3 < x < - 2 or x > 3$

Interval Notation:

$(- 3, - 2) \cup (3, \infty)$

Step 12