Solve for x (2x+1)(x-2)(3x-4)<=0

$(2 x + 1) (x - 2) (3 x - 4) \leq 0$

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

$2 x + 1 = 0$

$x - 2 = 0$

$3 x - 4 = 0$

Set $2 x + 1$ equal to $0$ and solve for $x$ .

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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 $3 x - 4$ equal to $0$ and solve for $x$ .

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Solve $3 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$3 x = 4$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Divide $x$ by $1$ .

$x = \frac{4}{3}$

The final solution is all the values that make $(2 x + 1) (x - 2) (3 x - 4) \leq 0$ true.

$x = - \frac{1}{2}, 2, \frac{4}{3}$

Use each root to create test intervals.

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

$- \frac{1}{2} < x < \frac{4}{3}$

$\frac{4}{3} < x < 2$

$x > 2$

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 $x < - \frac{1}{2}$ to see if it makes the inequality true.

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

$x = - 3$

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

$(2 (- 3) + 1) ((- 3) - 2) (3 (- 3) - 4) \leq 0$

The left side $- 325$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{1}{2} < x < \frac{4}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{1}{2} < x < \frac{4}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

$(2 (0) + 1) ((0) - 2) (3 (0) - 4) \leq 0$

The left side $8$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $\frac{4}{3} < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $\frac{4}{3} < x < 2$ and see if this value makes the original inequality true.

$x = 1.67$

Replace $x$ with $1.67$ in the original inequality.

$(2 (1.67) + 1) ((1.67) - 2) (3 (1.67) - 4) \leq 0$

The left side $- 1.446522$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

$(2 (4) + 1) ((4) - 2) (3 (4) - 4) \leq 0$

The left side $144$ is greater than the right side $0$ , which means that the given statement is false.

False

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

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

$- \frac{1}{2} < x < \frac{4}{3}$ False

$\frac{4}{3} < x < 2$ True

$x > 2$ False

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

$- \frac{1}{2} < x < \frac{4}{3}$ False

$\frac{4}{3} < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$x \leq - \frac{1}{2}$ or $\frac{4}{3} \leq x \leq 2$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{1}{2} or \frac{4}{3} \leq x \leq 2$

Interval Notation:

$(- \infty, - \frac{1}{2}] \cup [\frac{4}{3}, 2]$

Solve for x (2x+1)(x-2)(3x-4)<=0

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