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

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

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

$x + 3 = 0$

${(x - 4)}^{2} = 0$

$x^{2} - 4 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Add $4$ to both sides of the equation.

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

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

$x = - 3$

$x = 4$

$x = 2, - 2$

Consolidate the solutions.

$x = - 3, 4, 2, - 2$

Find the domain of $\frac{x + 3}{{(x - 4)}^{2} (x^{2} - 4)}$ .

Set the denominator in $\frac{x + 3}{{(x - 4)}^{2} (x^{2} - 4)}$ equal to $0$ to find where the expression is undefined.

${(x - 4)}^{2} (x^{2} - 4) = 0$

Solve for $x$ .

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

${(x - 4)}^{2} = 0$

$x^{2} - 4 = 0$

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 4)}^{2}$ equal to $0$ .

${(x - 4)}^{2} = 0$

Solve ${(x - 4)}^{2} = 0$ for $x$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x^{2} - 4$ equal to $0$ and solve for $x$ .

Set $x^{2} - 4$ equal to $0$ .

$x^{2} - 4 = 0$

Solve $x^{2} - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

The final solution is all the values that make ${(x - 4)}^{2} (x^{2} - 4) = 0$ true.

$x = 4, 2, - 2$

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

$(- \infty, - 2) \cup (- 2, 2) \cup (2, 4) \cup (4, \infty)$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 2$

$- 2 < x < 2$

$2 < x < 4$

$x > 4$

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 $x < - 3$ 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.

$\frac{(- 6) + 3}{{((- 6) - 4)}^{2} ({(- 6)}^{2} - 4)} \leq 0$

The left side $- 0.0009375$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = - 2.5$

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

$\frac{(- 2.5) + 3}{{((- 2.5) - 4)}^{2} ({(- 2.5)}^{2} - 4)} \leq 0$

The left side $0.00525969$ is greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

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

The left side $- 0.046875$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $2 < x < 4$ to see if it makes the inequality true.

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

$x = 3$

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

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

The left side $1.2$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

$\frac{(6) + 3}{{((6) - 4)}^{2} ({(6)}^{2} - 4)} \leq 0$

The left side $0.0703125$ 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.

$x < - 3$ True

$- 3 < x < - 2$ False

$- 2 < x < 2$ True

$2 < x < 4$ False

$x > 4$ False

$x < - 3$ True

$- 3 < x < - 2$ False

$- 2 < x < 2$ True

$2 < x < 4$ False

$x > 4$ False

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

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

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