Solve the Inequality for x 2x^2+24x<=-8x-120

$2 x^{2} + 24 x \leq - 8 x - 120$

Step 1

Move all terms containing $x$ to the left side of the inequality.

Add $8 x$ to both sides of the inequality.

$2 x^{2} + 24 x + 8 x \leq - 120$

Add $24 x$ and $8 x$ .

$2 x^{2} + 32 x \leq - 120$

Step 2

Convert the inequality to an equation.

$2 x^{2} + 32 x = - 120$

Step 3

Add $120$ to both sides of the equation.

$2 x^{2} + 32 x + 120 = 0$

Step 4

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 32 x + 120$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) + 32 x + 120 = 0$

Factor $2$ out of $32 x$ .

$2 (x^{2}) + 2 (16 x) + 120 = 0$

Factor $2$ out of $120$ .

$2 x^{2} + 2 (16 x) + 2 \cdot 60 = 0$

Factor $2$ out of $2 x^{2} + 2 (16 x)$ .

$2 (x^{2} + 16 x) + 2 \cdot 60 = 0$

Factor $2$ out of $2 (x^{2} + 16 x) + 2 \cdot 60$ .

$2 (x^{2} + 16 x + 60) = 0$

Factor.

Factor $x^{2} + 16 x + 60$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $60$ and whose sum is $16$ .

$6, 10$

Write the factored form using these integers.

$2 ((x + 6) (x + 10)) = 0$

Remove unnecessary parentheses.

$2 (x + 6) (x + 10) = 0$

Step 5

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

$x + 6 = 0$

$x + 10 = 0$

Step 6

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 7

Set $x + 10$ equal to $0$ and solve for $x$ .

Set $x + 10$ equal to $0$ .

$x + 10 = 0$

Subtract $10$ from both sides of the equation.

$x = - 10$

Step 8

The final solution is all the values that make $2 (x + 6) (x + 10) = 0$ true.

$x = - 6, - 10$

Step 9

Use each root to create test intervals.

$x < - 10$

$- 10 < x < - 6$

$x > - 6$

Step 10

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 $x < - 10$ to see if it makes the inequality true.

Choose a value on the interval $x < - 10$ and see if this value makes the original inequality true.

$x = - 12$

Replace $x$ with $- 12$ in the original inequality.

$2 {(- 12)}^{2} + 24 (- 12) \leq - 8 \cdot - 12 - 120$

The left side $0$ is greater than the right side $- 24$ , which means that the given statement is false.

False

Test a value on the interval $- 10 < x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $- 10 < x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$2 {(- 8)}^{2} + 24 (- 8) \leq - 8 \cdot - 8 - 120$

The left side $- 64$ is less than the right side $- 56$ , which means that the given statement is always true.

True

Test a value on the interval $x > - 6$ to see if it makes the inequality true.

Choose a value on the interval $x > - 6$ and see if this value makes the original inequality true.

$x = 0$

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

$2 {(0)}^{2} + 24 (0) \leq - 8 \cdot 0 - 120$

The left side $0$ is greater than the right side $- 120$ , which means that the given statement is false.

False

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

$x < - 10$ False

$- 10 < x < - 6$ True

$x > - 6$ False

$x < - 10$ False

$- 10 < x < - 6$ True

$x > - 6$ False

Step 11

The solution consists of all of the true intervals.

$- 10 \leq x \leq - 6$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$- 10 \leq x \leq - 6$

Interval Notation:

$[- 10, - 6]$

Step 13

Solve the Inequality for x 2x^2+24x<=-8x-120

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