Precalculus Examples

$2 x^{3} - 32 x < x^{2} - 16$

Step 1

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

$2 x^{3} - 32 x - x^{2} < - 16$

Step 2

Convert the inequality to an equation.

$2 x^{3} - 32 x - x^{2} = - 16$

Step 3

Add $16$ to both sides of the equation.

$2 x^{3} - 32 x - x^{2} + 16 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

Reorder terms.

$2 x^{3} - x^{2} - 32 x + 16 = 0$

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.

$(2 x^{3} - x^{2}) - 32 x + 16 = 0$

Step 4.2.2

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

$x^{2} (2 x - 1) - 16 (2 x - 1) = 0$

Step 4.3

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

$(2 x - 1) (x^{2} - 16) = 0$

Step 4.4

Rewrite $16$ as $4^{2}$ .

$(2 x - 1) (x^{2} - 4^{2}) = 0$

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 $b = 4$ .

$(2 x - 1) ((x + 4) (x - 4)) = 0$

Step 4.5.2

Remove unnecessary parentheses.

$(2 x - 1) (x + 4) (x - 4) = 0$

Step 5

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

$2 x - 1 = 0$

$x + 4 = 0$

$x - 4 = 0$

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 $x + 4$ equal to $0$ and solve for $x$ .

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Step 7.1

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

$x + 4 = 0$

Step 7.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 8

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

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Step 8.1

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

$x - 4 = 0$

Step 8.2

Add $4$ to both sides of the equation.

$x = 4$

Step 9

The final solution is all the values that make $(2 x - 1) (x + 4) (x - 4) = 0$ true.

$x = \frac{1}{2}, - 4, 4$

Step 10

Use each root to create test intervals.

$x < - 4$

$- 4 < x < \frac{1}{2}$

$\frac{1}{2} < x < 4$

$x > 4$

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 $x < - 4$ to see if it makes the inequality true.

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Step 11.1.1

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

$x = - 6$

Step 11.1.2

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

$2 {(- 6)}^{3} - 32 \cdot - 6 < {(- 6)}^{2} - 16$

Step 11.1.3

The left side $- 240$ is less than the right side $20$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 4 < x < \frac{1}{2}$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $- 4 < x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 11.2.2

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

$2 {(0)}^{3} - 32 \cdot 0 < {(0)}^{2} - 16$

Step 11.2.3

The left side $0$ is not less than the right side $- 16$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $\frac{1}{2} < x < 4$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $\frac{1}{2} < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 11.3.2

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

$2 {(2)}^{3} - 32 \cdot 2 < {(2)}^{2} - 16$

Step 11.3.3

The left side $- 48$ is less than the right side $- 12$ , which means that the given statement is always true.

True

Step 11.4

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

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Step 11.4.1

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

$x = 6$

Step 11.4.2

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

$2 {(6)}^{3} - 32 \cdot 6 < {(6)}^{2} - 16$

Step 11.4.3

The left side $240$ is not less than the right side $20$ , which means that the given statement is false.

False

Step 11.5

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

$x < - 4$ True

$- 4 < x < \frac{1}{2}$ False

$\frac{1}{2} < x < 4$ True

$x > 4$ False

$x < - 4$ True

$- 4 < x < \frac{1}{2}$ False

$\frac{1}{2} < x < 4$ True

$x > 4$ False

Step 12

The solution consists of all of the true intervals.

$x < - 4$ or $\frac{1}{2} < x < 4$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 4) \cup (\frac{1}{2}, 4)$

Step 14