Solve the Inequality for x (x-3)(x+1)^2(x-8)>=0

$(x - 3) {(x + 1)}^{2} (x - 8) \geq 0$

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

$x - 3 = 0$

${(x + 1)}^{2} = 0$

$x - 8 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set ${(x + 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 1)}^{2}$ equal to $0$ .

${(x + 1)}^{2} = 0$

Solve ${(x + 1)}^{2} = 0$ for $x$ .

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

The final solution is all the values that make $(x - 3) {(x + 1)}^{2} (x - 8) \geq 0$ true.

$x = 3, - 1, 8$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 3$

$3 < x < 8$

$x > 8$

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 $x < - 1$ to see if it makes the inequality true.

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

$x = - 4$

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

$((- 4) - 3) {((- 4) + 1)}^{2} ((- 4) - 8) \geq 0$

The left side $756$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) - 3) {((0) + 1)}^{2} ((0) - 8) \geq 0$

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

True

Test a value on the interval $3 < x < 8$ to see if it makes the inequality true.

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

$x = 6$

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

$((6) - 3) {((6) + 1)}^{2} ((6) - 8) \geq 0$

The left side $- 294$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

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

$((10) - 3) {((10) + 1)}^{2} ((10) - 8) \geq 0$

The left side $1694$ is greater 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.

$x < - 1$ True

$- 1 < x < 3$ True

$3 < x < 8$ False

$x > 8$ True

$x < - 1$ True

$- 1 < x < 3$ True

$3 < x < 8$ False

$x > 8$ True

The solution consists of all of the true intervals.

$x \leq - 1$ or $- 1 \leq x \leq 3$ or $x \geq 8$

Combine the intervals.

$x \leq 3 or x \geq 8$

The result can be shown in multiple forms.

Inequality Form:

$x \leq 3 or x \geq 8$

Interval Notation:

$(- \infty, 3] \cup [8, \infty)$

Solve the Inequality for x (x-3)(x+1)^2(x-8)>=0

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