Precalculus Examples

$2 x^{3} - 72 x < x^{2} - 36$

Step 1

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

$2 x^{3} - 72 x - x^{2} < - 36$

Step 2

Convert the inequality to an equation.

$2 x^{3} - 72 x - x^{2} = - 36$

Step 3

Add $36$ to both sides of the equation.

$2 x^{3} - 72 x - x^{2} + 36 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

Reorder terms.

$2 x^{3} - x^{2} - 72 x + 36 = 0$

Step 4.2

Factor out the greatest common factor from each group.

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Step 4.2.1

Group the first two terms and the last two terms.

$(2 x^{3} - x^{2}) - 72 x + 36 = 0$

Step 4.2.2

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

$x^{2} (2 x - 1) - 36 (2 x - 1) = 0$

Step 4.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x^{2} - 36) = 0$

Step 4.4

Rewrite $36$ as $6^{2}$ .

$(2 x - 1) (x^{2} - 6^{2}) = 0$

Step 4.5

Factor.

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Step 4.5.1

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 $b = 6$ .

$(2 x - 1) ((x + 6) (x - 6)) = 0$

Step 4.5.2

Remove unnecessary parentheses.

$(2 x - 1) (x + 6) (x - 6) = 0$

Step 5

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

$2 x - 1 = 0$

$x + 6 = 0$

$x - 6 = 0$

Step 6

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 6.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 6.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 6.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 6.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 7

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 7.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 8

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 8.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 8.2

Add $6$ to both sides of the equation.

$x = 6$

Step 9

The final solution is all the values that make $(2 x - 1) (x + 6) (x - 6) = 0$ true.

$x = \frac{1}{2}, - 6, 6$

Step 10

Use each root to create test intervals.

$x < - 6$

$- 6 < x < \frac{1}{2}$

$\frac{1}{2} < x < 6$

$x > 6$

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 $x < - 6$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

Step 11.1.2

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

$2 {(- 8)}^{3} - 72 \cdot - 8 < {(- 8)}^{2} - 36$

Step 11.1.3

The left side $- 448$ is less than the right side $28$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 6 < x < \frac{1}{2}$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $- 6 < x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 11.2.2

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

$2 {(0)}^{3} - 72 \cdot 0 < {(0)}^{2} - 36$

Step 11.2.3

The left side $0$ is not less than the right side $- 36$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $\frac{1}{2} < x < 6$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $\frac{1}{2} < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Step 11.3.2

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

$2 {(3)}^{3} - 72 \cdot 3 < {(3)}^{2} - 36$

Step 11.3.3

The left side $- 162$ is less than the right side $- 27$ , which means that the given statement is always true.

True

Step 11.4

Test a value on the interval $x > 6$ to see if it makes the inequality true.

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Step 11.4.1

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 11.4.2

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

$2 {(8)}^{3} - 72 \cdot 8 < {(8)}^{2} - 36$

Step 11.4.3

The left side $448$ is not less than the right side $28$ , which means that the given statement is false.

False

Step 11.5

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

$x < - 6$ True

$- 6 < x < \frac{1}{2}$ False

$\frac{1}{2} < x < 6$ True

$x > 6$ False

$x < - 6$ True

$- 6 < x < \frac{1}{2}$ False

$\frac{1}{2} < x < 6$ True

$x > 6$ False

Step 12

The solution consists of all of the true intervals.

$x < - 6$ or $\frac{1}{2} < x < 6$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 6) \cup (\frac{1}{2}, 6)$

Step 14