Precalculus Examples

$x^{3} - 2 x^{2} - 3 x < 0$

Step 1

Convert the inequality to an equation.

$x^{3} - 2 x^{2} - 3 x = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $x$ out of $x^{3} - 2 x^{2} - 3 x$ .

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Step 2.1.1

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 2 x^{2} - 3 x = 0$

Step 2.1.2

Factor $x$ out of $- 2 x^{2}$ .

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

Step 2.1.3

Factor $x$ out of $- 3 x$ .

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

Step 2.1.4

Factor $x$ out of $x \cdot x^{2} + x (- 2 x)$ .

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

Step 2.1.5

Factor $x$ out of $x (x^{2} - 2 x) + x \cdot - 3$ .

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

Step 2.2

Factor.

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Step 2.2.1

Factor $x^{2} - 2 x - 3$ using the AC method.

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Step 2.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 2.2.1.2

Write the factored form using these integers.

$x ((x - 3) (x + 1)) = 0$

Step 2.2.2

Remove unnecessary parentheses.

$x (x - 3) (x + 1) = 0$

Step 3

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

$x = 0$

$x - 3 = 0$

$x + 1 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

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

Add $3$ to both sides of the equation.

$x = 3$

Step 6

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

The final solution is all the values that make $x (x - 3) (x + 1) = 0$ true.

$x = 0, 3, - 1$

Step 8

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$0 < x < 3$

$x > 3$

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 $x < - 1$ to see if it makes the inequality true.

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Step 9.1.1

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

$x = - 4$

Step 9.1.2

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

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

Step 9.1.3

The left side $- 84$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 9.2.2

Replace $x$ with $- 0.5$ in the original inequality.

${(- 0.5)}^{3} - 2 {(- 0.5)}^{2} - 3 \cdot - 0.5 < 0$

Step 9.2.3

The left side $0.875$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.3

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

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Step 9.3.1

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

$x = 2$

Step 9.3.2

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

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

Step 9.3.3

The left side $- 6$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

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

Step 9.4.3

The left side $126$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.5

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

$x < - 1$ True

$- 1 < x < 0$ False

$0 < x < 3$ True

$x > 3$ False

$x < - 1$ True

$- 1 < x < 0$ False

$0 < x < 3$ True

$x > 3$ False

Step 10

The solution consists of all of the true intervals.

$x < - 1$ or $0 < x < 3$

Step 11

Convert the inequality to interval notation.

$(- \infty, - 1) \cup (0, 3)$

Step 12