Algebra Examples

$\frac{(x + 3) (x - 2)}{(x + 2) (x - 1)} \geq 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x + 3 = 0$

$x - 2 = 0$

$x + 2 = 0$

$x - 1 = 0$

Step 2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3

Add $2$ to both sides of the equation.

$x = 2$

Step 4

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

Add $1$ to both sides of the equation.

$x = 1$

Step 6

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = - 3$

$x = 2$

$x = - 2$

$x = 1$

Step 7

Consolidate the solutions.

$x = - 3, 2, - 2, 1$

Step 8

Find the domain of $\frac{(x + 3) (x - 2)}{(x + 2) (x - 1)}$ .

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Step 8.1

Set the denominator in $\frac{(x + 3) (x - 2)}{(x + 2) (x - 1)}$ equal to $0$ to find where the expression is undefined.

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

Step 8.2

Solve for $x$ .

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Step 8.2.1

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

$x + 2 = 0$

$x - 1 = 0$

Step 8.2.2

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

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Step 8.2.2.1

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

$x + 2 = 0$

Step 8.2.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 8.2.3

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

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Step 8.2.3.1

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

$x - 1 = 0$

Step 8.2.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 8.2.4

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

$x = - 2, 1$

Step 8.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 2) \cup (- 2, 1) \cup (1, \infty)$

Step 9

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 2$

$- 2 < x < 1$

$1 < x < 2$

$x > 2$

Step 10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 10.1

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Step 10.1.2

Replace $x$ with $- 6$ in the original inequality.

$\frac{((- 6) + 3) ((- 6) - 2)}{((- 6) + 2) ((- 6) - 1)} \geq 0$

Step 10.1.3

The left side $0.\overline{857142}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 10.2

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

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Step 10.2.1

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

$x = - 2.5$

Step 10.2.2

Replace $x$ with $- 2.5$ in the original inequality.

$\frac{((- 2.5) + 3) ((- 2.5) - 2)}{((- 2.5) + 2) ((- 2.5) - 1)} \geq 0$

Step 10.2.3

The left side $- 1.\overline{285714}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 10.3

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

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Step 10.3.1

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

$x = 0$

Step 10.3.2

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

$\frac{((0) + 3) ((0) - 2)}{((0) + 2) ((0) - 1)} \geq 0$

Step 10.3.3

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 10.4

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

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Step 10.4.1

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

$x = 1.5$

Step 10.4.2

Replace $x$ with $1.5$ in the original inequality.

$\frac{((1.5) + 3) ((1.5) - 2)}{((1.5) + 2) ((1.5) - 1)} \geq 0$

Step 10.4.3

The left side $- 1.\overline{285714}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 10.5

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 10.5.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 10.5.2

Replace $x$ with $4$ in the original inequality.

$\frac{((4) + 3) ((4) - 2)}{((4) + 2) ((4) - 1)} \geq 0$

Step 10.5.3

The left side $0.\overline{7}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 10.6

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

$x < - 3$ True

$- 3 < x < - 2$ False

$- 2 < x < 1$ True

$1 < x < 2$ False

$x > 2$ True

$x < - 3$ True

$- 3 < x < - 2$ False

$- 2 < x < 1$ True

$1 < x < 2$ False

$x > 2$ True

Step 11

The solution consists of all of the true intervals.

$x \leq - 3$ or $- 2 < x < 1$ or $x \geq 2$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or - 2 < x < 1 or x \geq 2$

Interval Notation:

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

Step 13