Solve for x x^3-16x<=0

$x^{3} - 16 x \leq 0$

Step 1

Convert the inequality to an equation.

$x^{3} - 16 x = 0$

Step 2

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 16 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 16 x = 0$

Factor $x$ out of $- 16 x$ .

$x \cdot x^{2} + x \cdot - 16 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 16$ .

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

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

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

Factor.

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 $b = 4$ .

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

Remove unnecessary parentheses.

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

Step 3

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

$x = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 6

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 7

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

$x = 0, - 4, 4$

Step 8

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 0$

$0 < x < 4$

$x > 4$

Step 9

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 $x < - 4$ 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.

${(- 6)}^{3} - 16 \cdot - 6 \leq 0$

The left side $- 120$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 4 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

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

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

The left side $24$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

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

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

The left side $- 24$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

${(6)}^{3} - 16 \cdot 6 \leq 0$

The left side $120$ is greater than the right side $0$ , which means that the given statement is false.

False

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

$x < - 4$ True

$- 4 < x < 0$ False

$0 < x < 4$ True

$x > 4$ False

$x < - 4$ True

$- 4 < x < 0$ False

$0 < x < 4$ True

$x > 4$ False

Step 10

The solution consists of all of the true intervals.

$x \leq - 4$ or $0 \leq x \leq 4$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 4 or 0 \leq x \leq 4$

Interval Notation:

$(- \infty, - 4] \cup [0, 4]$

Step 12

Solve for x x^3-16x<=0

Related Questions

Support LCMGCF.com by disabling your adblocker.