Precalculus Examples

$x^{3} + 8 x > 6 x^{2}$

Step 1

Subtract $6 x^{2}$ from both sides of the inequality.

$x^{3} + 8 x - 6 x^{2} > 0$

Step 2

Convert the inequality to an equation.

$x^{3} + 8 x - 6 x^{2} = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Factor $x$ out of $x^{3} + 8 x - 6 x^{2}$ .

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Step 3.1.1

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

$x \cdot x^{2} + 8 x - 6 x^{2} = 0$

Step 3.1.2

Factor $x$ out of $8 x$ .

$x \cdot x^{2} + x \cdot 8 - 6 x^{2} = 0$

Step 3.1.3

Factor $x$ out of $- 6 x^{2}$ .

$x \cdot x^{2} + x \cdot 8 + x (- 6 x) = 0$

Step 3.1.4

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

$x (x^{2} + 8) + x (- 6 x) = 0$

Step 3.1.5

Factor $x$ out of $x (x^{2} + 8) + x (- 6 x)$ .

$x (x^{2} + 8 - 6 x) = 0$

Step 3.2

Factor.

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Step 3.2.1

Factor $x^{2} + 8 - 6 x$ 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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Step 3.2.1.2

Write the factored form using these integers.

$x ((x - 4) (x - 2)) = 0$

Step 3.2.2

Remove unnecessary parentheses.

$x (x - 4) (x - 2) = 0$

Step 4

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

$x = 0$

$x - 4 = 0$

$x - 2 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 6.2

Add $4$ to both sides of the equation.

$x = 4$

Step 7

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

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Step 7.1

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

$x - 2 = 0$

Step 7.2

Add $2$ to both sides of the equation.

$x = 2$

Step 8

The final solution is all the values that make $x (x - 4) (x - 2) = 0$ true.

$x = 0, 4, 2$

Step 9

Use each root to create test intervals.

$x < 0$

$0 < x < 2$

$2 < x < 4$

$x > 4$

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.

${(- 2)}^{3} + 8 (- 2) > 6 {(- 2)}^{2}$

Step 10.1.3

The left side $- 24$ is not greater than the right side $24$ , which means that the given statement is false.

False

Step 10.2

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

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Step 10.2.1

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

$x = 1$

Step 10.2.2

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

${(1)}^{3} + 8 (1) > 6 {(1)}^{2}$

Step 10.2.3

The left side $9$ is greater than the right side $6$ , which means that the given statement is always true.

True

Step 10.3

Test a value on the interval $2 < x < 4$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $2 < x < 4$ and see if this value makes the original inequality true.

$x = 3$

Step 10.3.2

Replace $x$ with $3$ in the original inequality.

${(3)}^{3} + 8 (3) > 6 {(3)}^{2}$

Step 10.3.3

The left side $51$ is not greater than the right side $54$ , which means that the given statement is false.

False

Step 10.4

Test a value on the interval $x > 4$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Step 10.4.2

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

${(6)}^{3} + 8 (6) > 6 {(6)}^{2}$

Step 10.4.3

The left side $264$ is greater than the right side $216$ , which means that the given statement is always true.

True

Step 10.5

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

$x < 0$ False

$0 < x < 2$ True

$2 < x < 4$ False

$x > 4$ True

$x < 0$ False

$0 < x < 2$ True

$2 < x < 4$ False

$x > 4$ True

Step 11

The solution consists of all of the true intervals.

$0 < x < 2$ or $x > 4$

Step 12

Convert the inequality to interval notation.

$(0, 2) \cup (4, \infty)$

Step 13