Convert to Set Notation (3x+1)(x-8)>=0

$(3 x + 1) (x - 8) \geq 0$

Step 1

Solve $x \leq - \frac{1}{3} or x \geq 8$ .

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

$3 x + 1 = 0$

$x - 8 = 0$

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 $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

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

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

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

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

Use each root to create test intervals.

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

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

$x > 8$

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

$(3 (- 3) + 1) ((- 3) - 8) \geq 0$

The left side $88$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$(3 (0) + 1) ((0) - 8) \geq 0$

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

False

Test a value on the interval $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$(3 (10) + 1) ((10) - 8) \geq 0$

The left side $62$ 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.

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

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

$x > 8$ True

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

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

$x > 8$ True

The solution consists of all of the true intervals.

$x \leq - \frac{1}{3}$ or $x \geq 8$

Step 2

Use the inequality $x \leq - \frac{1}{3} or x \geq 8$ to build the set notation.

${x | x \leq - \frac{1}{3} or x \geq 8}$

Step 3

Convert to Set Notation (3x+1)(x-8)>=0

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