Convert to Set Notation (x+3)(x+1)>0

$(x + 3) (x + 1) > 0$

Solve $x < - 3 or x > - 1$ .

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

$x + 3 = 0$

$x + 1 = 0$

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.

$x = - 3$

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 $(x + 3) (x + 1) > 0$ true.

$x = - 3, - 1$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$x > - 1$

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.

$((- 6) + 3) ((- 6) + 1) > 0$

The left side $15$ is greater 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.

$((- 2) + 3) ((- 2) + 1) > 0$

The left side $- 1$ is not greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$((0) + 3) ((0) + 1) > 0$

The left side $3$ 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.

$x < - 3$ True

$- 3 < x < - 1$ False

$x > - 1$ True

$x < - 3$ True

$- 3 < x < - 1$ False

$x > - 1$ True

The solution consists of all of the true intervals.

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

Use the inequality $x < - 3 or x > - 1$ to build the set notation.

${x | x < - 3 or x > - 1}$

Convert to Set Notation (x+3)(x+1)>0

Related Questions

Support LCMGCF.com by disabling your adblocker.