Solve the Inequality for x (x+6)(x^2-16)(1-x)<=0

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

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 $x$ .

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

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 3

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

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

$x^{2} - 16 = 0$

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

Add $16$ to both sides of the equation.

$x^{2} = 16$

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

$x = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

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

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

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

$x = \pm 4$

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

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

$x = 4$

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

$x = - 4$

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

$x = 4, - 4$

Step 4

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

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

$1 - x = 0$

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

Subtract $1$ from both sides of the equation.

$- x = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

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.

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

Step 6

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 4$

$- 4 < x < 1$

$1 < x < 4$

$x > 4$

Step 7

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = - 8$

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

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

The left side $- 864$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 6 < x < - 4$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < - 4$ and see if this value makes the original inequality true.

$x = - 5$

Replace $x$ with $- 5$ in the original inequality.

$((- 5) + 6) ({(- 5)}^{2} - 16) (1 - (- 5)) \leq 0$

The left side $54$ is greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

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

The left side $- 96$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 4$ to see if it makes the inequality true.

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

$x = 2$

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

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

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

False

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

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

$x = 6$

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

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

The left side $- 1200$ is less than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 6$ True

$- 6 < x < - 4$ False

$- 4 < x < 1$ True

$1 < x < 4$ False

$x > 4$ True

$x < - 6$ True

$- 6 < x < - 4$ False

$- 4 < x < 1$ True

$1 < x < 4$ False

$x > 4$ True

Step 8

The solution consists of all of the true intervals.

$x \leq - 6$ or $- 4 \leq x \leq 1$ or $x \geq 4$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 6 or - 4 \leq x \leq 1 or x \geq 4$

Interval Notation:

$(- \infty, - 6] \cup [- 4, 1] \cup [4, \infty)$

Step 10

Solve the Inequality for x (x+6)(x^2-16)(1-x)<=0

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