Solve the Inequality for x x^3+3x^2-x-3>=0

$x^{3} + 3 x^{2} - x - 3 \geq 0$

Step 1

Convert the inequality to an equation.

$x^{3} + 3 x^{2} - x - 3 = 0$

Step 2

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

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

Rewrite $1$ as $1^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$(x + 3) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x + 3) (x + 1) (x - 1) = 0$

Step 3

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

$x + 3 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 4

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.

$x = - 3$

Step 5

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.

$x = - 1$

Step 6

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 7

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

$x = - 3, - 1, 1$

Step 8

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 1$

$x > 1$

Step 9

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.

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

The left side $- 105$ is less than the right side $0$ , which means that the given statement is false.

False

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.

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

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

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

The left side $- 3$ is less 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.

$x = 4$

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

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

The left side $105$ 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.

$x < - 3$ False

$- 3 < x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

$x < - 3$ False

$- 3 < x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

Step 10

The solution consists of all of the true intervals.

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

Step 11

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 12

Solve the Inequality for x x^3+3x^2-x-3>=0

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