Precalculus Examples

$(x + 6) (x^{2} + 16) (1 - x) \leq 0$

Step 1

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

$x + 6 = 0$

$x^{2} + 16 = 0$

$1 - x = 0$

Step 2

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

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Step 2.1

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

$x + 6 = 0$

Step 2.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 3

Set $x^{2} + 16$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x^{2} + 16$ equal to $0$ .

$x^{2} + 16 = 0$

Step 3.2

Solve $x^{2} + 16 = 0$ for $x$ .

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Step 3.2.1

Subtract $16$ from both sides of the equation.

$x^{2} = - 16$

Step 3.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 16}$

Step 3.2.3

Simplify $\pm \sqrt{- 16}$ .

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Step 3.2.3.1

Rewrite $- 16$ as $- 1 (16)$ .

$x = \pm \sqrt{- 1 (16)}$

Step 3.2.3.2

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{16}$

Step 3.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{16}$

Step 3.2.3.4

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

$x = \pm i \cdot \sqrt{4^{2}}$

Step 3.2.3.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 4$

Step 3.2.3.6

Move $4$ to the left of $i$ .

$x = \pm 4 i$

Step 3.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 4 i$

Step 3.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 4 i$

Step 3.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 4 i, - 4 i$

Step 4

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

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Step 4.1

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

$1 - x = 0$

Step 4.2

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

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 4.2.2

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

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Step 4.2.2.1

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

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

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 4.2.2.2.2

Divide $x$ by $1$ .

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

Step 4.2.2.3

Simplify the right side.

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Step 4.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 5

The final solution is all the values that make $(x + 6) (x^{2} + 16) (1 - x) \leq 0$ true.

$x = - 6, 4 i, - 4 i, 1$

Step 6

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 1$

$x > 1$

Step 7

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 7.1

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

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Step 7.1.1

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

$x = - 8$

Step 7.1.2

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

$((- 8) + 6) ({(- 8)}^{2} + 16) (1 - (- 8)) \leq 0$

Step 7.1.3

The left side $- 1440$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 7.2

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

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Step 7.2.1

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

$x = 0$

Step 7.2.2

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

$((0) + 6) ({(0)}^{2} + 16) (1 - (0)) \leq 0$

Step 7.2.3

The left side $96$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 7.3

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

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Step 7.3.1

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

$x = 4$

Step 7.3.2

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

$((4) + 6) ({(4)}^{2} + 16) (1 - (4)) \leq 0$

Step 7.3.3

The left side $- 960$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 7.4

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

$x < - 6$ True

$- 6 < x < 1$ False

$x > 1$ True

$x < - 6$ True

$- 6 < x < 1$ False

$x > 1$ True

Step 8

The solution consists of all of the true intervals.

$x \leq - 6$ or $x \geq 1$

Step 9

Convert the inequality to interval notation.

$(- \infty, - 6] \cup [1, \infty)$

Step 10