Solve the Inequality for x 3x^2+5x-2>=0

$3 x^{2} + 5 x - 2 \geq 0$

Convert the inequality to an equation.

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

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 $b = 5$ .

Factor $5$ out of $5 x$ .

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

Rewrite $5$ as $- 1$ plus $6$

$3 x^{2} + (- 1 + 6) x - 2 = 0$

Apply the distributive property.

$3 x^{2} - 1 x + 6 x - 2 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 1 x) + 6 x - 2 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

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

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

$3 x - 1 = 0$

$x + 2 = 0$

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

Use each root to create test intervals.

$x < - 2$

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

$x > \frac{1}{3}$

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 $x < - 2$ 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.

$3 {(- 4)}^{2} + 5 (- 4) - 2 \geq 0$

The left side $26$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$3 {(0)}^{2} + 5 (0) - 2 \geq 0$

The left side $- 2$ is less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 3$

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

$3 {(3)}^{2} + 5 (3) - 2 \geq 0$

The left side $40$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 2$ True

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

$x > \frac{1}{3}$ True

$x < - 2$ True

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

$x > \frac{1}{3}$ True

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq \frac{1}{3}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or x \geq \frac{1}{3}$

Interval Notation:

$(- \infty, - 2] \cup [\frac{1}{3}, \infty)$

Solve the Inequality for x 3x^2+5x-2>=0

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