Solve the Inequality for x x^3+4x^2>=4x+16

$x^{3} + 4 x^{2} \geq 4 x + 16$

Step 1

Subtract $4 x$ from both sides of the inequality.

$x^{3} + 4 x^{2} - 4 x \geq 16$

Step 2

Convert the inequality to an equation.

$x^{3} + 4 x^{2} - 4 x = 16$

Step 3

Subtract $16$ from both sides of the equation.

$x^{3} + 4 x^{2} - 4 x - 16 = 0$

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.

$(x^{3} + 4 x^{2}) - 4 x - 16 = 0$

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

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

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

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

Rewrite $4$ as $2^{2}$ .

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

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 $b = 2$ .

$(x + 4) ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x + 4) (x + 2) (x - 2) = 0$

Step 5

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

$x + 4 = 0$

$x + 2 = 0$

$x - 2 = 0$

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 $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 8

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 9

The final solution is all the values that make $(x + 4) (x + 2) (x - 2) = 0$ true.

$x = - 4, - 2, 2$

Step 10

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - 2$

$- 2 < x < 2$

$x > 2$

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.

${(- 6)}^{3} + 4 {(- 6)}^{2} \geq 4 (- 6) + 16$

The left side $- 72$ is less than the right side $- 8$ , which means that the given statement is false.

False

Test a value on the interval $- 4 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < - 2$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

${(- 3)}^{3} + 4 {(- 3)}^{2} \geq 4 (- 3) + 16$

The left side $9$ is greater than the right side $4$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{3} + 4 {(0)}^{2} \geq 4 (0) + 16$

The left side $0$ is less than the right side $16$ , which means that the given statement is false.

False

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

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

${(4)}^{3} + 4 {(4)}^{2} \geq 4 (4) + 16$

The left side $128$ is greater than the right side $32$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 4$ False

$- 4 < x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

$x < - 4$ False

$- 4 < x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

Step 12

The solution consists of all of the true intervals.

$- 4 \leq x \leq - 2$ or $x \geq 2$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$- 4 \leq x \leq - 2 or x \geq 2$

Interval Notation:

$[- 4, - 2] \cup [2, \infty)$

Step 14

Solve the Inequality for x x^3+4x^2>=4x+16

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