Solve the Inequality for x -3x^2+3x+4<=-2

$- 3 x^{2} + 3 x + 4 \leq - 2$

Convert the inequality to an equation.

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

Add $2$ to both sides of the equation.

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

Add $4$ and $2$ .

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

Factor the left side of the equation.

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

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

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

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

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

Factor $- 3$ out of $6$ .

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

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

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

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

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

Factor.

Factor $x^{2} - x - 2$ 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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$- 3 ((x - 2) (x + 1)) = 0$

Remove unnecessary parentheses.

$- 3 (x - 2) (x + 1) = 0$

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

$x - 2 = 0$

$x + 1 = 0$

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.

$x = 2$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- 3 (x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 2$

$x > 2$

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 $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

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

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

The left side $- 56$ is less than the right side $- 2$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$- 3 {(0)}^{2} + 3 (0) + 4 \leq - 2$

The left side $4$ is greater than the right side $- 2$ , 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.

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

The left side $- 32$ is less than the right side $- 2$ , which means that the given statement is always true.

True

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

$x < - 1$ True

$- 1 < x < 2$ False

$x > 2$ True

$x < - 1$ True

$- 1 < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$x \leq - 1$ or $x \geq 2$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 1 or x \geq 2$

Interval Notation:

$(- \infty, - 1] \cup [2, \infty)$

Solve the Inequality for x -3x^2+3x+4<=-2

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