Solve the Inequality for x -3x^2+6>x+4

$- 3 x^{2} + 6 > x + 4$

Subtract $x$ from both sides of the inequality.

$- 3 x^{2} + 6 - x > 4$

Convert the inequality to an equation.

$- 3 x^{2} + 6 - x = 4$

Subtract $4$ from both sides of the equation.

$- 3 x^{2} + 6 - x - 4 = 0$

Subtract $4$ from $6$ .

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

Factor the left side of the equation.

Factor $- 1$ out of $- 3 x^{2} - x + 2$ .

Factor $- 1$ out of $- 3 x^{2}$ .

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

Factor $- 1$ out of $- x$ .

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

Rewrite $2$ as $- 1 (- 2)$ .

$- (3 x^{2}) - (x) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (3 x^{2}) - (x)$ .

$- (3 x^{2} + x) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (3 x^{2} + x) - 1 (- 2)$ .

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

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 2 = - 6$ and whose sum is $b = 1$ .

Multiply by $1$ .

$- (3 x^{2} + 1 x - 2) = 0$

Rewrite $1$ as $- 2$ plus $3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 2$ .

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

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 $0$ .

$3 x - 2 = 0$

$x + 1 = 0$

Set $3 x - 2$ equal to $0$ and solve for $x$ .

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Solve $3 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

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.

$x = \frac{2}{3}, - 1$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < \frac{2}{3}$

$x > \frac{2}{3}$

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.

$- 3 {(- 4)}^{2} + 6 > (- 4) + 4$

The left side $- 42$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < \frac{2}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < \frac{2}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

$- 3 {(0)}^{2} + 6 > (0) + 4$

The left side $6$ is greater than the right side $4$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{2}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{2}{3}$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$- 3 {(3)}^{2} + 6 > (3) + 4$

The left side $- 21$ is not greater than the right side $7$ , which means that the given statement is false.

False

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

$x < - 1$ False

$- 1 < x < \frac{2}{3}$ True

$x > \frac{2}{3}$ False

$x < - 1$ False

$- 1 < x < \frac{2}{3}$ True

$x > \frac{2}{3}$ False

The solution consists of all of the true intervals.

$- 1 < x < \frac{2}{3}$

The result can be shown in multiple forms.

Inequality Form:

$- 1 < x < \frac{2}{3}$

Interval Notation:

$(- 1, \frac{2}{3})$

Solve the Inequality for x -3x^2+6>x+4

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