Convert to Interval Notation (x^2+x-12)/(x+1)>=0

$\frac{x^{2} + x - 12}{x + 1} \geq 0$

Step 1

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

$x^{2} + x - 12 = 0$

$x + 1 = 0$

Step 2

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$(x - 3) (x + 4) = 0$

Step 3

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

$x - 3 = 0$

$x + 4 = 0$

Step 4

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.

$x = 3$

Step 5

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 6

The final solution is all the values that make $(x - 3) (x + 4) = 0$ true.

$x = 3, - 4$

Step 7

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8

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

$x = 3, - 4$

$x = - 1$

Step 9

Consolidate the solutions.

$x = 3, - 4, - 1$

Step 10

Find the domain of $\frac{x^{2} + x - 12}{x + 1}$ .

Set the denominator in $\frac{x^{2} + x - 12}{x + 1}$ equal to $0$ to find where the expression is undefined.

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

Step 11

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - 1$

$- 1 < x < 3$

$x > 3$

Step 12

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 $x < - 4$ 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.

$\frac{{(- 6)}^{2} - 6 - 12}{(- 6) + 1} \geq 0$

The left side $- 3.6$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 4 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < - 1$ and see if this value makes the original inequality true.

$x = - 2$

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

$\frac{{(- 2)}^{2} - 2 - 12}{(- 2) + 1} \geq 0$

The left side $10$ 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.

$\frac{{(0)}^{2} + 0 - 12}{(0) + 1} \geq 0$

The left side $- 12$ is 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.

$\frac{{(6)}^{2} + 6 - 12}{(6) + 1} \geq 0$

The left side $4.\overline{285714}$ 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.

$x < - 4$ False

$- 4 < x < - 1$ True

$- 1 < x < 3$ False

$x > 3$ True

$x < - 4$ False

$- 4 < x < - 1$ True

$- 1 < x < 3$ False

$x > 3$ True

Step 13

The solution consists of all of the true intervals.

$- 4 \leq x < - 1$ or $x \geq 3$

Step 14

Convert the inequality to interval notation.

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

Step 15

Convert to Interval Notation (x^2+x-12)/(x+1)>=0

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