Solve the Inequality for x ((x-3)(x+2))/(x-1)<=0

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

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

$x - 3 = 0$

$x + 2 = 0$

$x - 1 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Subtract $2$ from both sides of the equation.

$x = - 2$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 3$

$x = - 2$

$x = 1$

Consolidate the solutions.

$x = 3, - 2, 1$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 1$

$1 < x < 3$

$x > 3$

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.

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

The left side $- 2.8$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

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

The left side $6$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

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

The left side $- 4$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 6$

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

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

The left side $4.8$ 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.

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Solve the Inequality for x ((x-3)(x+2))/(x-1)<=0

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