Solve the Inequality for x -3x^2-9x>-3x-240

$- 3 x^{2} - 9 x > - 3 x - 240$

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

Add $3 x$ to both sides of the inequality.

$- 3 x^{2} - 9 x + 3 x > - 240$

Add $- 9 x$ and $3 x$ .

$- 3 x^{2} - 6 x > - 240$

Convert the inequality to an equation.

$- 3 x^{2} - 6 x = - 240$

Add $240$ to both sides of the equation.

$- 3 x^{2} - 6 x + 240 = 0$

Factor the left side of the equation.

Factor $- 3$ out of $- 3 x^{2} - 6 x + 240$ .

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

$- 3 x^{2} - 6 x + 240 = 0$

Factor $- 3$ out of $- 6 x$ .

$- 3 x^{2} - 3 (2 x) + 240 = 0$

Factor $- 3$ out of $240$ .

$- 3 x^{2} - 3 (2 x) - 3 \cdot - 80 = 0$

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

$- 3 (x^{2} + 2 x) - 3 \cdot - 80 = 0$

Factor $- 3$ out of $- 3 (x^{2} + 2 x) - 3 (- 80)$ .

$- 3 (x^{2} + 2 x - 80) = 0$

Factor.

Factor $x^{2} + 2 x - 80$ 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 $- 80$ and whose sum is $2$ .

$- 8, 10$

Write the factored form using these integers.

$- 3 ((x - 8) (x + 10)) = 0$

Remove unnecessary parentheses.

$- 3 (x - 8) (x + 10) = 0$

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

$x - 8 = 0$

$x + 10 = 0$

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.

$x = 8$

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.

$x = - 10$

The final solution is all the values that make $- 3 (x - 8) (x + 10) = 0$ true.

$x = 8, - 10$

Use each root to create test intervals.

$x < - 10$

$- 10 < x < 8$

$x > 8$

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.

$- 3 {(- 12)}^{2} - 9 \cdot - 12 > - 3 \cdot - 12 - 240$

The left side $- 324$ is not greater than the right side $- 204$ , which means that the given statement is false.

False

Test a value on the interval $- 10 < x < 8$ to see if it makes the inequality true.

Choose a value on the interval $- 10 < x < 8$ and see if this value makes the original inequality true.

$x = 0$

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

$- 3 {(0)}^{2} - 9 \cdot 0 > - 3 \cdot 0 - 240$

The left side $0$ is greater than the right side $- 240$ , which means that the given statement is always true.

True

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.

$- 3 {(10)}^{2} - 9 \cdot 10 > - 3 \cdot 10 - 240$

The left side $- 390$ is not greater than the right side $- 270$ , which means that the given statement is false.

False

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

$x < - 10$ False

$- 10 < x < 8$ True

$x > 8$ False

$x < - 10$ False

$- 10 < x < 8$ True

$x > 8$ False

The solution consists of all of the true intervals.

$- 10 < x < 8$

The result can be shown in multiple forms.

Inequality Form:

$- 10 < x < 8$

Interval Notation:

$(- 10, 8)$

Solve the Inequality for x -3x^2-9x>-3x-240

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