Precalculus Examples

$\frac{x^{2} + 9 x + 8}{x^{2}} \div \frac{x^{2} - 1}{x^{3} + 3 x^{2} - 4 x}$

Step 1

Set the denominator in $\frac{x^{2} + 9 x + 8}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$x = \pm \sqrt{0}$

Step 2.2

Simplify $\pm \sqrt{0}$ .

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Step 2.2.1

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

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

Step 2.2.2

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

$x = \pm 0$

Step 2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 3

Set the denominator in $\frac{x^{2} - 1}{x^{3} + 3 x^{2} - 4 x}$ equal to $0$ to find where the expression is undefined.

$x^{3} + 3 x^{2} - 4 x = 0$

Step 4

Solve for $x$ .

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Step 4.1

Factor the left side of the equation.

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Step 4.1.1

Factor $x$ out of $x^{3} + 3 x^{2} - 4 x$ .

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Step 4.1.1.1

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} + 3 x^{2} - 4 x = 0$

Step 4.1.1.2

Factor $x$ out of $3 x^{2}$ .

$x \cdot x^{2} + x (3 x) - 4 x = 0$

Step 4.1.1.3

Factor $x$ out of $- 4 x$ .

$x \cdot x^{2} + x (3 x) + x \cdot - 4 = 0$

Step 4.1.1.4

Factor $x$ out of $x \cdot x^{2} + x (3 x)$ .

$x (x^{2} + 3 x) + x \cdot - 4 = 0$

Step 4.1.1.5

Factor $x$ out of $x (x^{2} + 3 x) + x \cdot - 4$ .

$x (x^{2} + 3 x - 4) = 0$

Step 4.1.2

Factor.

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Step 4.1.2.1

Factor $x^{2} + 3 x - 4$ using the AC method.

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Step 4.1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $3$ .

$- 1, 4$

Step 4.1.2.1.2

Write the factored form using these integers.

$x ((x - 1) (x + 4)) = 0$

Step 4.1.2.2

Remove unnecessary parentheses.

$x (x - 1) (x + 4) = 0$

Step 4.2

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

$x = 0$

$x - 1 = 0$

$x + 4 = 0$

Step 4.3

Set $x$ equal to $0$ .

$x = 0$

Step 4.4

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

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Step 4.4.1

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

$x - 1 = 0$

Step 4.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 4.5

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 4.5.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 4.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 4.6

The final solution is all the values that make $x (x - 1) (x + 4) = 0$ true.

$x = 0, 1, - 4$

Step 5

Set the denominator in $\frac{x^{2} + 9 x + 8}{x^{2}} \div \frac{x^{2} - 1}{x^{3} + 3 x^{2} - 4 x}$ equal to $0$ to find where the expression is undefined.

$\frac{x^{2} - 1}{x^{3} + 3 x^{2} - 4 x} = 0$

Step 6

Solve for $x$ .

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Step 6.1

Set the numerator equal to zero.

$x^{2} - 1 = 0$

Step 6.2

Solve the equation for $x$ .

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Step 6.2.1

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 6.2.2

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

$x = \pm \sqrt{1}$

Step 6.2.3

Any root of $1$ is $1$ .

$x = \pm 1$

Step 6.2.4

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

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Step 6.2.4.1

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

$x = 1$

Step 6.2.4.2

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

$x = - 1$

Step 6.2.4.3

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

$x = 1, - 1$

Step 6.3

Exclude the solutions that do not make $\frac{x^{2} - 1}{x^{3} + 3 x^{2} - 4 x} = 0$ true.

$x = - 1$

Step 7

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

Interval Notation:

$(- \infty, - 4) \cup (- 4, - 1) \cup (- 1, 0) \cup (0, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x \neq 0, 1, - 1, - 4}$

Step 8