Solve the Inequality for x x^3+4x^2>x+4

$x^{3} + 4 x^{2} > x + 4$

Step 1

Subtract $x$ from both sides of the inequality.

$x^{3} + 4 x^{2} - x > 4$

Step 2

Convert the inequality to an equation.

$x^{3} + 4 x^{2} - x = 4$

Step 3

Subtract $4$ from both sides of the equation.

$x^{3} + 4 x^{2} - x - 4 = 0$

Step 4

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.

$(x^{3} + 4 x^{2}) - x - 4 = 0$

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

$x^{2} (x + 4) - (x + 4) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 4$ .

$(x + 4) (x^{2} - 1) = 0$

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

$(x + 4) (x^{2} - 1^{2}) = 0$

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 $b = 1$ .

$(x + 4) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x + 4) (x + 1) (x - 1) = 0$

Step 5

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

$x + 4 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 6

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 7

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 8

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 9

The final solution is all the values that make $(x + 4) (x + 1) (x - 1) = 0$ true.

$x = - 4, - 1, 1$

Step 10

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - 1$

$- 1 < x < 1$

$x > 1$

Step 11

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.

${(- 6)}^{3} + 4 {(- 6)}^{2} > (- 6) + 4$

The left side $- 72$ is not greater than the right side $- 2$ , 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.

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

The left side $8$ is greater than the right side $2$ , 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.

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

The left side $0$ is not greater than the right side $4$ , 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.

${(4)}^{3} + 4 {(4)}^{2} > (4) + 4$

The left side $128$ is greater than the right side $8$ , 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

$- 1 < x < 1$ False

$x > 1$ True

$x < - 4$ False

$- 4 < x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

Step 12

The solution consists of all of the true intervals.

$- 4 < x < - 1$ or $x > 1$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$- 4 < x < - 1 or x > 1$

Interval Notation:

$(- 4, - 1) \cup (1, \infty)$

Step 14

Solve the Inequality for x x^3+4x^2>x+4

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