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

$(3 - x) (x + 1) (x + 5) < 0$

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

$3 - x = 0$

$x + 1 = 0$

$x + 5 = 0$

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

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

$3 - x = 0$

Solve $3 - x = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

Set $x + 1$ equal to $0$ and solve for $x$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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.

$x = - 5$

The final solution is all the values that make $(3 - x) (x + 1) (x + 5) < 0$ true.

$x = 3, - 1, - 5$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 1$

$- 1 < x < 3$

$x > 3$

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.

$(3 - (- 8)) ((- 8) + 1) ((- 8) + 5) < 0$

The left side $231$ is not less 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.

$(3 - (- 3)) ((- 3) + 1) ((- 3) + 5) < 0$

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 $- 1 < x < 3$ 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.

$(3 - (0)) ((0) + 1) ((0) + 5) < 0$

The left side $15$ is not less than the right side $0$ , which means that the given statement is false.

False

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.

$(3 - (6)) ((6) + 1) ((6) + 5) < 0$

The left side $- 231$ 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.

$x < - 5$ False

$- 5 < x < - 1$ True

$- 1 < x < 3$ False

$x > 3$ True

$x < - 5$ False

$- 5 < x < - 1$ True

$- 1 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$- 5 < x < - 1$ or $x > 3$

The result can be shown in multiple forms.

Inequality Form:

$- 5 < x < - 1 or x > 3$

Interval Notation:

$(- 5, - 1) \cup (3, \infty)$

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

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