Solve for x (x+1)(x-3)^2>0

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

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

$x + 1 = 0$

${(x - 3)}^{2} = 0$

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 ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 3)}^{2}$ equal to $0$ .

${(x - 3)}^{2} = 0$

Solve ${(x - 3)}^{2} = 0$ for $x$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = - 1, 3$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 3$

$x > 3$

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 $x < - 1$ 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.

$x = - 4$

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

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

The left side $- 147$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

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

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

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.

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

The left side $63$ 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.

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ True

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ True

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Solve for x (x+1)(x-3)^2>0

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