Solve the Inequality for x (15-x)(3x+8)(x-5)^3<=0

$(15 - x) (3 x + 8) {(x - 5)}^{3} \leq 0$

Step 1

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

$15 - x = 0$

$3 x + 8 = 0$

${(x - 5)}^{3} = 0$

Step 2

Set $15 - x$ equal to $0$ and solve for $x$ .

Set $15 - x$ equal to $0$ .

$15 - x = 0$

Solve $15 - x = 0$ for $x$ .

Subtract $15$ from both sides of the equation.

$- x = - 15$

Divide each term in $- x = - 15$ by $- 1$ and simplify.

Divide each term in $- x = - 15$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 15}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 15}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 15}{- 1}$

Simplify the right side.

Divide $- 15$ by $- 1$ .

$x = 15$

Step 3

Set $3 x + 8$ equal to $0$ and solve for $x$ .

Set $3 x + 8$ equal to $0$ .

$3 x + 8 = 0$

Solve $3 x + 8 = 0$ for $x$ .

Subtract $8$ from both sides of the equation.

$3 x = - 8$

Divide each term in $3 x = - 8$ by $3$ and simplify.

Divide each term in $3 x = - 8$ by $3$ .

$\frac{3 x}{3} = \frac{- 8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 8}{3}$

Divide $x$ by $1$ .

$x = \frac{- 8}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{8}{3}$

Step 4

Set ${(x - 5)}^{3}$ equal to $0$ and solve for $x$ .

Set ${(x - 5)}^{3}$ equal to $0$ .

${(x - 5)}^{3} = 0$

Solve ${(x - 5)}^{3} = 0$ for $x$ .

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 5

The final solution is all the values that make $(15 - x) (3 x + 8) {(x - 5)}^{3} \leq 0$ true.

$x = 15, - \frac{8}{3}, 5$

Step 6

Use each root to create test intervals.

$x < - \frac{8}{3}$

$- \frac{8}{3} < x < 5$

$5 < x < 15$

$x > 15$

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 $x < - \frac{8}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{8}{3}$ and see if this value makes the original inequality true.

$x = - 5$

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

$(15 - (- 5)) (3 (- 5) + 8) {((- 5) - 5)}^{3} \leq 0$

The left side $140000$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{8}{3} < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{8}{3} < x < 5$ and see if this value makes the original inequality true.

$x = 0$

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

$(15 - (0)) (3 (0) + 8) {((0) - 5)}^{3} \leq 0$

The left side $- 15000$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $5 < x < 15$ to see if it makes the inequality true.

Choose a value on the interval $5 < x < 15$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$(15 - (10)) (3 (10) + 8) {((10) - 5)}^{3} \leq 0$

The left side $23750$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 15$ to see if it makes the inequality true.

Choose a value on the interval $x > 15$ and see if this value makes the original inequality true.

$x = 18$

Replace $x$ with $18$ in the original inequality.

$(15 - (18)) (3 (18) + 8) {((18) - 5)}^{3} \leq 0$

The left side $- 408642$ 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.

$x < - \frac{8}{3}$ False

$- \frac{8}{3} < x < 5$ True

$5 < x < 15$ False

$x > 15$ True

$x < - \frac{8}{3}$ False

$- \frac{8}{3} < x < 5$ True

$5 < x < 15$ False

$x > 15$ True

Step 8

The solution consists of all of the true intervals.

$- \frac{8}{3} \leq x \leq 5$ or $x \geq 15$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$- \frac{8}{3} \leq x \leq 5 or x \geq 15$

Interval Notation:

$[- \frac{8}{3}, 5] \cup [15, \infty)$

Step 10

Solve the Inequality for x (15-x)(3x+8)(x-5)^3<=0

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