Solve the Inequality for x ((x-5)(x+1))/(x-3)<=0

$\frac{(x - 5) (x + 1)}{x - 3} \leq 0$

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

$x - 5 = 0$

$x + 1 = 0$

$x - 3 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Subtract $1$ from both sides of the equation.

$x = - 1$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 5$

$x = - 1$

$x = 3$

Consolidate the solutions.

$x = 5, - 1, 3$

Find the domain of $\frac{(x - 5) (x + 1)}{x - 3}$ .

Set the denominator in $\frac{(x - 5) (x + 1)}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

$(- \infty, 3) \cup (3, \infty)$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 3$

$3 < x < 5$

$x > 5$

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 $x < - 1$ to see if it makes the inequality true.

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

$x = - 4$

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

$\frac{((- 4) - 5) ((- 4) + 1)}{(- 4) - 3} \leq 0$

The left side $- 3.\overline{857142}$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

$x = 0$

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

$\frac{((0) - 5) ((0) + 1)}{(0) - 3} \leq 0$

The left side $1.\overline{6}$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $3 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{((4) - 5) ((4) + 1)}{(4) - 3} \leq 0$

The left side $- 5$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$\frac{((8) - 5) ((8) + 1)}{(8) - 3} \leq 0$

The left side $5.4$ 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.

$x < - 1$ True

$- 1 < x < 3$ False

$3 < x < 5$ True

$x > 5$ False

$x < - 1$ True

$- 1 < x < 3$ False

$3 < x < 5$ True

$x > 5$ False

The solution consists of all of the true intervals.

$x \leq - 1$ or $3 < x \leq 5$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 1 or 3 < x \leq 5$

Interval Notation:

$(- \infty, - 1] \cup (3, 5]$

Solve the Inequality for x ((x-5)(x+1))/(x-3)<=0

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