Precalculus Examples

$x^{3} - 8 x^{2} \leq - 15 x$

Step 1

Add $15 x$ to both sides of the inequality.

$x^{3} - 8 x^{2} + 15 x \leq 0$

Step 2

Convert the inequality to an equation.

$x^{3} - 8 x^{2} + 15 x = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Factor $x$ out of $x^{3} - 8 x^{2} + 15 x$ .

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Step 3.1.1

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

$x \cdot x^{2} - 8 x^{2} + 15 x = 0$

Step 3.1.2

Factor $x$ out of $- 8 x^{2}$ .

$x \cdot x^{2} + x (- 8 x) + 15 x = 0$

Step 3.1.3

Factor $x$ out of $15 x$ .

$x \cdot x^{2} + x (- 8 x) + x \cdot 15 = 0$

Step 3.1.4

Factor $x$ out of $x \cdot x^{2} + x (- 8 x)$ .

$x (x^{2} - 8 x) + x \cdot 15 = 0$

Step 3.1.5

Factor $x$ out of $x (x^{2} - 8 x) + x \cdot 15$ .

$x (x^{2} - 8 x + 15) = 0$

Step 3.2

Factor.

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Step 3.2.1

Factor $x^{2} - 8 x + 15$ using the AC method.

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Step 3.2.1.1

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 $15$ and whose sum is $- 8$ .

$- 5, - 3$

Step 3.2.1.2

Write the factored form using these integers.

$x ((x - 5) (x - 3)) = 0$

Step 3.2.2

Remove unnecessary parentheses.

$x (x - 5) (x - 3) = 0$

Step 4

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

$x = 0$

$x - 5 = 0$

$x - 3 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 6.1

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

$x - 5 = 0$

Step 6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 7

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

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Step 7.1

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

$x - 3 = 0$

Step 7.2

Add $3$ to both sides of the equation.

$x = 3$

Step 8

The final solution is all the values that make $x (x - 5) (x - 3) = 0$ true.

$x = 0, 5, 3$

Step 9

Use each root to create test intervals.

$x < 0$

$0 < x < 3$

$3 < x < 5$

$x > 5$

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 $x < 0$ to see if it makes the inequality true.

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Step 10.1.1

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

$x = - 2$

Step 10.1.2

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

${(- 2)}^{3} - 8 {(- 2)}^{2} \leq - 15 \cdot - 2$

Step 10.1.3

The left side $- 40$ is less than the right side $30$ , which means that the given statement is always true.

True

Step 10.2

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

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Step 10.2.1

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

$x = 2$

Step 10.2.2

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

${(2)}^{3} - 8 {(2)}^{2} \leq - 15 \cdot 2$

Step 10.2.3

The left side $- 24$ is greater than the right side $- 30$ , which means that the given statement is false.

False

Step 10.3

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

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Step 10.3.1

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

$x = 4$

Step 10.3.2

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

${(4)}^{3} - 8 {(4)}^{2} \leq - 15 \cdot 4$

Step 10.3.3

The left side $- 64$ is less than the right side $- 60$ , which means that the given statement is always true.

True

Step 10.4

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

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Step 10.4.1

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

$x = 8$

Step 10.4.2

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

${(8)}^{3} - 8 {(8)}^{2} \leq - 15 \cdot 8$

Step 10.4.3

The left side $0$ is greater than the right side $- 120$ , which means that the given statement is false.

False

Step 10.5

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

$x < 0$ True

$0 < x < 3$ False

$3 < x < 5$ True

$x > 5$ False

$x < 0$ True

$0 < x < 3$ False

$3 < x < 5$ True

$x > 5$ False

Step 11

The solution consists of all of the true intervals.

$x \leq 0$ or $3 \leq x \leq 5$

Step 12

Convert the inequality to interval notation.

$(- \infty, 0] \cup [3, 5]$

Step 13