Solve the Inequality for x 3x^2+10x-99>-2x-3

$3 x^{2} + 10 x - 99 > - 2 x - 3$

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

Add $2 x$ to both sides of the inequality.

$3 x^{2} + 10 x - 99 + 2 x > - 3$

Add $10 x$ and $2 x$ .

$3 x^{2} + 12 x - 99 > - 3$

Convert the inequality to an equation.

$3 x^{2} + 12 x - 99 = - 3$

Add $3$ to both sides of the equation.

$3 x^{2} + 12 x - 99 + 3 = 0$

Add $- 99$ and $3$ .

$3 x^{2} + 12 x - 96 = 0$

Factor the left side of the equation.

Factor $3$ out of $3 x^{2} + 12 x - 96$ .

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

$3 (x^{2}) + 12 x - 96 = 0$

Factor $3$ out of $12 x$ .

$3 (x^{2}) + 3 (4 x) - 96 = 0$

Factor $3$ out of $- 96$ .

$3 x^{2} + 3 (4 x) + 3 \cdot - 32 = 0$

Factor $3$ out of $3 x^{2} + 3 (4 x)$ .

$3 (x^{2} + 4 x) + 3 \cdot - 32 = 0$

Factor $3$ out of $3 (x^{2} + 4 x) + 3 \cdot - 32$ .

$3 (x^{2} + 4 x - 32) = 0$

Factor.

Factor $x^{2} + 4 x - 32$ 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 $- 32$ and whose sum is $4$ .

$- 4, 8$

Write the factored form using these integers.

$3 ((x - 4) (x + 8)) = 0$

Remove unnecessary parentheses.

$3 (x - 4) (x + 8) = 0$

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

$x - 4 = 0$

$x + 8 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

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

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $3 (x - 4) (x + 8) = 0$ true.

$x = 4, - 8$

Use each root to create test intervals.

$x < - 8$

$- 8 < x < 4$

$x > 4$

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 $x < - 8$ 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.

$3 {(- 10)}^{2} + 10 (- 10) - 99 > - 2 \cdot - 10 - 3$

The left side $101$ is greater than the right side $17$ , which means that the given statement is always true.

True

Test a value on the interval $- 8 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 8 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

$3 {(0)}^{2} + 10 (0) - 99 > - 2 \cdot 0 - 3$

The left side $- 99$ is not greater than the right side $- 3$ , which means that the given statement is false.

False

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.

$3 {(6)}^{2} + 10 (6) - 99 > - 2 \cdot 6 - 3$

The left side $69$ is greater than the right side $- 15$ , which means that the given statement is always true.

True

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

$x < - 8$ True

$- 8 < x < 4$ False

$x > 4$ True

$x < - 8$ True

$- 8 < x < 4$ False

$x > 4$ True

The solution consists of all of the true intervals.

$x < - 8$ or $x > 4$

The result can be shown in multiple forms.

Inequality Form:

$x < - 8 or x > 4$

Interval Notation:

$(- \infty, - 8) \cup (4, \infty)$

Solve the Inequality for x 3x^2+10x-99>-2x-3

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