Solve the Inequality for x 3x^2-8x-120>x

$3 x^{2} - 8 x - 120 > x$

Step 1

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

Subtract $x$ from both sides of the inequality.

$3 x^{2} - 8 x - 120 - x > 0$

Subtract $x$ from $- 8 x$ .

$3 x^{2} - 9 x - 120 > 0$

Step 2

Convert the inequality to an equation.

$3 x^{2} - 9 x - 120 = 0$

Step 3

Factor the left side of the equation.

Factor $3$ out of $3 x^{2} - 9 x - 120$ .

Factor $3$ out of $3 x^{2}$ .

$3 (x^{2}) - 9 x - 120 = 0$

Factor $3$ out of $- 9 x$ .

$3 (x^{2}) + 3 (- 3 x) - 120 = 0$

Factor $3$ out of $- 120$ .

$3 x^{2} + 3 (- 3 x) + 3 \cdot - 40 = 0$

Factor $3$ out of $3 x^{2} + 3 (- 3 x)$ .

$3 (x^{2} - 3 x) + 3 \cdot - 40 = 0$

Factor $3$ out of $3 (x^{2} - 3 x) + 3 \cdot - 40$ .

$3 (x^{2} - 3 x - 40) = 0$

Factor.

Factor $x^{2} - 3 x - 40$ 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 $- 40$ and whose sum is $- 3$ .

$- 8, 5$

Write the factored form using these integers.

$3 ((x - 8) (x + 5)) = 0$

Remove unnecessary parentheses.

$3 (x - 8) (x + 5) = 0$

Step 4

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

$x - 8 = 0$

$x + 5 = 0$

Step 5

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Step 6

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 7

The final solution is all the values that make $3 (x - 8) (x + 5) = 0$ true.

$x = 8, - 5$

Step 8

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 8$

$x > 8$

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 $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

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

$3 {(- 8)}^{2} - 8 \cdot - 8 - 120 > - 8$

The left side $136$ is greater than the right side $- 8$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < 8$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 8$ and see if this value makes the original inequality true.

$x = 0$

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

$3 {(0)}^{2} - 8 \cdot 0 - 120 > 0$

The left side $- 120$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$3 {(10)}^{2} - 8 \cdot 10 - 120 > 10$

The left side $100$ is greater than the right side $10$ , which means that the given statement is always true.

True

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

$x < - 5$ True

$- 5 < x < 8$ False

$x > 8$ True

$x < - 5$ True

$- 5 < x < 8$ False

$x > 8$ True

Step 10

The solution consists of all of the true intervals.

$x < - 5$ or $x > 8$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x < - 5 or x > 8$

Interval Notation:

$(- \infty, - 5) \cup (8, \infty)$

Step 12

Solve the Inequality for x 3x^2-8x-120>x

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