Algebra Examples

$\frac{x^{2} {(x + 1)}^{3}}{(x - 7) {(x + 3)}^{2} (- x^{2} - 1)} \leq 0$

Step 1

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

$x^{2} = 0$

${(x + 1)}^{3} = 0$

$x - 7 = 0$

${(x + 3)}^{2} = 0$

$- x^{2} - 1 = 0$

Step 2

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

$x = \pm \sqrt{0}$

Step 3

Simplify $\pm \sqrt{0}$ .

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Step 3.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 3.2

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

$x = \pm 0$

Step 3.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 4

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

Add $7$ to both sides of the equation.

$x = 7$

Step 7

Set the $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 8

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 9

Add $1$ to both sides of the equation.

$- x^{2} = 1$

Step 10

Divide each term in $- x^{2} = 1$ by $- 1$ and simplify.

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Step 10.1

Divide each term in $- x^{2} = 1$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{1}{- 1}$

Step 10.2

Simplify the left side.

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Step 10.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{1}{- 1}$

Step 10.2.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{1}{- 1}$

Step 10.3

Simplify the right side.

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Step 10.3.1

Divide $1$ by $- 1$ .

$x^{2} = - 1$

Step 11

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

$x = \pm \sqrt{- 1}$

Step 12

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 13

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

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Step 13.1

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

$x = i$

Step 13.2

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

$x = - i$

Step 13.3

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

$x = i, - i$

Step 14

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

$x = 0$

$x = - 1$

$x = 7$

$x = - 3$

$x = i, - i$

Step 15

Consolidate the solutions.

$x = 0, - 1, 7, - 3$

Step 16

Find the domain of $\frac{x^{2} {(x + 1)}^{3}}{(x - 7) {(x + 3)}^{2} (- x^{2} - 1)}$ .

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Step 16.1

Set the denominator in $\frac{x^{2} {(x + 1)}^{3}}{(x - 7) {(x + 3)}^{2} (- x^{2} - 1)}$ equal to $0$ to find where the expression is undefined.

$(x - 7) {(x + 3)}^{2} (- x^{2} - 1) = 0$

Step 16.2

Solve for $x$ .

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Step 16.2.1

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

$x - 7 = 0$

${(x + 3)}^{2} = 0$

$- x^{2} - 1 = 0$

Step 16.2.2

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 16.2.2.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 16.2.2.2

Add $7$ to both sides of the equation.

$x = 7$

Step 16.2.3

Set ${(x + 3)}^{2}$ equal to $0$ and solve for $x$ .

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Step 16.2.3.1

Set ${(x + 3)}^{2}$ equal to $0$ .

${(x + 3)}^{2} = 0$

Step 16.2.3.2

Solve ${(x + 3)}^{2} = 0$ for $x$ .

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Step 16.2.3.2.1

Set the $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 16.2.3.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 16.2.4

Set $- x^{2} - 1$ equal to $0$ and solve for $x$ .

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Step 16.2.4.1

Set $- x^{2} - 1$ equal to $0$ .

$- x^{2} - 1 = 0$

Step 16.2.4.2

Solve $- x^{2} - 1 = 0$ for $x$ .

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Step 16.2.4.2.1

Add $1$ to both sides of the equation.

$- x^{2} = 1$

Step 16.2.4.2.2

Divide each term in $- x^{2} = 1$ by $- 1$ and simplify.

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Step 16.2.4.2.2.1

Divide each term in $- x^{2} = 1$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{1}{- 1}$

Step 16.2.4.2.2.2

Simplify the left side.

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Step 16.2.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{1}{- 1}$

Step 16.2.4.2.2.2.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{1}{- 1}$

Step 16.2.4.2.2.3

Simplify the right side.

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Step 16.2.4.2.2.3.1

Divide $1$ by $- 1$ .

$x^{2} = - 1$

Step 16.2.4.2.3

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

$x = \pm \sqrt{- 1}$

Step 16.2.4.2.4

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 16.2.4.2.5

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

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Step 16.2.4.2.5.1

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

$x = i$

Step 16.2.4.2.5.2

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

$x = - i$

Step 16.2.4.2.5.3

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

$x = i, - i$

Step 16.2.5

The final solution is all the values that make $(x - 7) {(x + 3)}^{2} (- x^{2} - 1) = 0$ true.

$x = 7, - 3, i, - i$

Step 16.3

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

$(- \infty, - 3) \cup (- 3, 7) \cup (7, \infty)$

Step 17

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 0$

$0 < x < 7$

$x > 7$

Step 18

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 18.1

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

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Step 18.1.1

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

$x = - 6$

Step 18.1.2

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

$\frac{{(- 6)}^{2} {((- 6) + 1)}^{3}}{((- 6) - 7) {((- 6) + 3)}^{2} (- {(- 6)}^{2} - 1)} \leq 0$

Step 18.1.3

The left side $- 1.\overline{039501}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 18.2

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

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Step 18.2.1

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

$x = - 2$

Step 18.2.2

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

$\frac{{(- 2)}^{2} {((- 2) + 1)}^{3}}{((- 2) - 7) {((- 2) + 3)}^{2} (- {(- 2)}^{2} - 1)} \leq 0$

Step 18.2.3

The left side $- 0.0\overline{8}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 18.3

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

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Step 18.3.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 18.3.2

Replace $x$ with $- 0.5$ in the original inequality.

$\frac{{(- 0.5)}^{2} {((- 0.5) + 1)}^{3}}{((- 0.5) - 7) {((- 0.5) + 3)}^{2} (- {(- 0.5)}^{2} - 1)} \leq 0$

Step 18.3.3

The left side $0.0005\overline{3}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 18.4

Test a value on the interval $0 < x < 7$ to see if it makes the inequality true.

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Step 18.4.1

Choose a value on the interval $0 < x < 7$ and see if this value makes the original inequality true.

$x = 4$

Step 18.4.2

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

$\frac{{(4)}^{2} {((4) + 1)}^{3}}{((4) - 7) {((4) + 3)}^{2} (- {(4)}^{2} - 1)} \leq 0$

Step 18.4.3

The left side $0.80032012$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 18.5

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 18.5.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 18.5.2

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

$\frac{{(10)}^{2} {((10) + 1)}^{3}}{((10) - 7) {((10) + 3)}^{2} (- {(10)}^{2} - 1)} \leq 0$

Step 18.5.3

The left side $- 2.599254$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 18.6

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

$x < - 3$ True

$- 3 < x < - 1$ True

$- 1 < x < 0$ False

$0 < x < 7$ False

$x > 7$ True

$x < - 3$ True

$- 3 < x < - 1$ True

$- 1 < x < 0$ False

$0 < x < 7$ False

$x > 7$ True

Step 19

The solution consists of all of the true intervals.

$x < - 3$ or $- 3 < x \leq - 1$ or $x > 7$ or $x = 0$

Step 20

Combine the intervals.

$x < - 3 or - 3 < x \leq - 1 or x = 0 or x > 7$

Step 21

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or - 3 < x \leq - 1 or x = 0 or x > 7$

Interval Notation:

$(- \infty, - 3) \cup (- 3, - 1] \cup [0, 0] \cup (7, \infty)$

Step 22