Solve the Inequality for x (x^2-1)(x+5)(x-3)<=0

$(x^{2} - 1) (x + 5) (x - 3) \leq 0$

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

$x^{2} - 1 = 0$

$x + 5 = 0$

$x - 3 = 0$

Set $x^{2} - 1$ equal to $0$ and solve for $x$ .

Set $x^{2} - 1$ equal to $0$ .

$x^{2} - 1 = 0$

Solve $x^{2} - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$x^{2} = 1$

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

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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.

$x = 3$

The final solution is all the values that make $(x^{2} - 1) (x + 5) (x - 3) \leq 0$ true.

$x = 1, - 1, - 5, 3$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 1$

$- 1 < x < 1$

$1 < x < 3$

$x > 3$

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 $x < - 5$ to see if it makes the inequality true.

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

$x = - 8$

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

$({(- 8)}^{2} - 1) ((- 8) + 5) ((- 8) - 3) \leq 0$

The left side $2079$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < - 1$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

$({(- 3)}^{2} - 1) ((- 3) + 5) ((- 3) - 3) \leq 0$

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

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

$x = 0$

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

$({(0)}^{2} - 1) ((0) + 5) ((0) - 3) \leq 0$

The left side $15$ is greater than the right side $0$ , which means that the given statement is false.

False

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.

$x = 2$

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

$({(2)}^{2} - 1) ((2) + 5) ((2) - 3) \leq 0$

The left side $- 21$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 6$

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

$({(6)}^{2} - 1) ((6) + 5) ((6) - 3) \leq 0$

The left side $1155$ 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.

$x < - 5$ False

$- 5 < x < - 1$ True

$- 1 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

$x < - 5$ False

$- 5 < x < - 1$ True

$- 1 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 5 \leq x \leq - 1$ or $1 \leq x \leq 3$

The result can be shown in multiple forms.

Inequality Form:

$- 5 \leq x \leq - 1 or 1 \leq x \leq 3$

Interval Notation:

$[- 5, - 1] \cup [1, 3]$

Solve the Inequality for x (x^2-1)(x+5)(x-3)<=0

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