Solve the Inequality for x (4-x)(x+3)(x^2-1)<0

$(4 - x) (x + 3) (x^{2} - 1) < 0$

Step 1

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

$4 - x = 0$

$x + 3 = 0$

$x^{2} - 1 = 0$

Step 2

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

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

$4 - x = 0$

Solve $4 - x = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$- x = - 4$

Divide each term in $- x = - 4$ by $- 1$ and simplify.

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x = 4$

Step 3

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 4

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.

$x = 1, - 1$

Step 5

The final solution is all the values that make $(4 - x) (x + 3) (x^{2} - 1) < 0$ true.

$x = 4, - 3, 1, - 1$

Step 6

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 1$

$1 < x < 4$

$x > 4$

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 $x < - 3$ 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.

$(4 - (- 6)) ((- 6) + 3) ({(- 6)}^{2} - 1) < 0$

The left side $- 1050$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = - 2$

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

$(4 - (- 2)) ((- 2) + 3) ({(- 2)}^{2} - 1) < 0$

The left side $18$ is not less than the right side $0$ , which means that the given statement is false.

False

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.

$(4 - (0)) ((0) + 3) ({(0)}^{2} - 1) < 0$

The left side $- 12$ 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.

$(4 - (2)) ((2) + 3) ({(2)}^{2} - 1) < 0$

The left side $30$ is not less 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.

$(4 - (6)) ((6) + 3) ({(6)}^{2} - 1) < 0$

The left side $- 630$ 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.

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 4$ False

$x > 4$ True

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 4$ False

$x > 4$ True

Step 8

The solution consists of all of the true intervals.

$x < - 3$ or $- 1 < x < 1$ or $x > 4$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or - 1 < x < 1 or x > 4$

Interval Notation:

$(- \infty, - 3) \cup (- 1, 1) \cup (4, \infty)$

Step 10

Solve the Inequality for x (4-x)(x+3)(x^2-1)<0

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