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

$\frac{x^{2} - x - 2}{x + 3} < 0$

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x^{2} - x - 2 = 0$

$x + 3 = 0$

Factor $x^{2} - x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$(x - 2) (x + 1) = 0$

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

$x - 2 = 0$

$x + 1 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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.

$x = - 1$

The final solution is all the values that make $(x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

Subtract $3$ from both sides of the equation.

$x = - 3$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 2, - 1$

$x = - 3$

Consolidate the solutions.

$x = 2, - 1, - 3$

Find the domain of $\frac{x^{2} - x - 2}{x + 3}$ .

Set the denominator in $\frac{x^{2} - x - 2}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The domain is all values of $x$ that make the expression defined.

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

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 2$

$x > 2$

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.

$\frac{{(- 6)}^{2} - (- 6) - 2}{(- 6) + 3} < 0$

The left side $- 13.\overline{3}$ 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.

$\frac{{(- 2)}^{2} - (- 2) - 2}{(- 2) + 3} < 0$

The left side $4$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{{(0)}^{2} - (0) - 2}{(0) + 3} < 0$

The left side $- 0.\overline{6}$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

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

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

The left side $1.\overline{428571}$ is not less 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.

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 2$ True

$x > 2$ False

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$x < - 3$ or $- 1 < x < 2$

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or - 1 < x < 2$

Interval Notation:

$(- \infty, - 3) \cup (- 1, 2)$

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

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