Solve the Inequality for x -3x^2+9x+8<4x

$- 3 x^{2} + 9 x + 8 < 4 x$

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

Subtract $4 x$ from both sides of the inequality.

$- 3 x^{2} + 9 x + 8 - 4 x < 0$

Subtract $4 x$ from $9 x$ .

$- 3 x^{2} + 5 x + 8 < 0$

Convert the inequality to an equation.

$- 3 x^{2} + 5 x + 8 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 3 x^{2} + 5 x + 8$ .

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

$- (3 x^{2}) + 5 x + 8 = 0$

Factor $- 1$ out of $5 x$ .

$- (3 x^{2}) - (- 5 x) + 8 = 0$

Rewrite $8$ as $- 1 (- 8)$ .

$- (3 x^{2}) - (- 5 x) - 1 \cdot - 8 = 0$

Factor $- 1$ out of $- (3 x^{2}) - (- 5 x)$ .

$- (3 x^{2} - 5 x) - 1 \cdot - 8 = 0$

Factor $- 1$ out of $- (3 x^{2} - 5 x) - 1 (- 8)$ .

$- (3 x^{2} - 5 x - 8) = 0$

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 $a \cdot c = 3 \cdot - 8 = - 24$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$- (3 x^{2} - 5 x - 8) = 0$

Rewrite $- 5$ as $3$ plus $- 8$

$- (3 x^{2} + (3 - 8) x - 8) = 0$

Apply the distributive property.

$- (3 x^{2} + 3 x - 8 x - 8) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((3 x^{2} + 3 x) - 8 x - 8) = 0$

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

$- (3 x (x + 1) - 8 (x + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$- ((x + 1) (3 x - 8)) = 0$

Remove unnecessary parentheses.

$- (x + 1) (3 x - 8) = 0$

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

$x + 1 = 0$

$3 x - 8 = 0$

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.

$x = - 1$

Set $3 x - 8$ equal to $0$ and solve for $x$ .

Set $3 x - 8$ equal to $0$ .

$3 x - 8 = 0$

Solve $3 x - 8 = 0$ for $x$ .

Add $8$ to both sides of the equation.

$3 x = 8$

Divide each term in $3 x = 8$ by $3$ and simplify.

Divide each term in $3 x = 8$ by $3$ .

$\frac{3 x}{3} = \frac{8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{8}{3}$

Divide $x$ by $1$ .

$x = \frac{8}{3}$

The final solution is all the values that make $- (x + 1) (3 x - 8) = 0$ true.

$x = - 1, \frac{8}{3}$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < \frac{8}{3}$

$x > \frac{8}{3}$

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.

$- 3 {(- 4)}^{2} + 9 (- 4) + 8 < 4 (- 4)$

The left side $- 76$ is less than the right side $- 16$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < \frac{8}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < \frac{8}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

$- 3 {(0)}^{2} + 9 (0) + 8 < 4 (0)$

The left side $8$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{8}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{8}{3}$ and see if this value makes the original inequality true.

$x = 5$

Replace $x$ with $5$ in the original inequality.

$- 3 {(5)}^{2} + 9 (5) + 8 < 4 (5)$

The left side $- 22$ is less than the right side $20$ , which means that the given statement is always true.

True

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

$x < - 1$ True

$- 1 < x < \frac{8}{3}$ False

$x > \frac{8}{3}$ True

$x < - 1$ True

$- 1 < x < \frac{8}{3}$ False

$x > \frac{8}{3}$ True

The solution consists of all of the true intervals.

$x < - 1$ or $x > \frac{8}{3}$

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or x > \frac{8}{3}$

Interval Notation:

$(- \infty, - 1) \cup (\frac{8}{3}, \infty)$

Solve the Inequality for x -3x^2+9x+8<4x

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