Precalculus Examples

$\frac{x^{4} + x^{3} - 30 x^{2}}{x^{2} - 3 x - 18} \div \frac{x^{3} + x^{2} - 30 x}{x^{2} - 36}$

Step 1

Set the denominator in $\frac{x^{4} + x^{3} - 30 x^{2}}{x^{2} - 3 x - 18}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 3 x - 18 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x^{2} - 3 x - 18$ using the AC method.

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Step 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 $- 18$ and whose sum is $- 3$ .

$- 6, 3$

Step 2.1.2

Write the factored form using these integers.

$(x - 6) (x + 3) = 0$

Step 2.2

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

$x - 6 = 0$

$x + 3 = 0$

Step 2.3

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 2.3.2

Add $6$ to both sides of the equation.

$x = 6$

Step 2.4

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.4.1

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

$x + 3 = 0$

Step 2.4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.5

The final solution is all the values that make $(x - 6) (x + 3) = 0$ true.

$x = 6, - 3$

Step 3

Set the denominator in $\frac{x^{3} + x^{2} - 30 x}{x^{2} - 36}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 36 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Add $36$ to both sides of the equation.

$x^{2} = 36$

Step 4.2

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

$x = \pm \sqrt{36}$

Step 4.3

Simplify $\pm \sqrt{36}$ .

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Step 4.3.1

Rewrite $36$ as $6^{2}$ .

$x = \pm \sqrt{6^{2}}$

Step 4.3.2

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

$x = \pm 6$

Step 4.4

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

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Step 4.4.1

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

$x = 6$

Step 4.4.2

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

$x = - 6$

Step 4.4.3

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

$x = 6, - 6$

Step 5

Set the denominator in $\frac{x^{4} + x^{3} - 30 x^{2}}{x^{2} - 3 x - 18} \div \frac{x^{3} + x^{2} - 30 x}{x^{2} - 36}$ equal to $0$ to find where the expression is undefined.

$\frac{x^{3} + x^{2} - 30 x}{x^{2} - 36} = 0$

Step 6

Solve for $x$ .

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Step 6.1

Set the numerator equal to zero.

$x^{3} + x^{2} - 30 x = 0$

Step 6.2

Solve the equation for $x$ .

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Step 6.2.1

Factor the left side of the equation.

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Step 6.2.1.1

Factor $x$ out of $x^{3} + x^{2} - 30 x$ .

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Step 6.2.1.1.1

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

$x \cdot x^{2} + x^{2} - 30 x = 0$

Step 6.2.1.1.2

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

$x \cdot x^{2} + x \cdot x - 30 x = 0$

Step 6.2.1.1.3

Factor $x$ out of $- 30 x$ .

$x \cdot x^{2} + x \cdot x + x \cdot - 30 = 0$

Step 6.2.1.1.4

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

$x (x^{2} + x) + x \cdot - 30 = 0$

Step 6.2.1.1.5

Factor $x$ out of $x (x^{2} + x) + x \cdot - 30$ .

$x (x^{2} + x - 30) = 0$

Step 6.2.1.2

Factor.

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Step 6.2.1.2.1

Factor $x^{2} + x - 30$ using the AC method.

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Step 6.2.1.2.1.1

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 $- 30$ and whose sum is $1$ .

$- 5, 6$

Step 6.2.1.2.1.2

Write the factored form using these integers.

$x ((x - 5) (x + 6)) = 0$

Step 6.2.1.2.2

Remove unnecessary parentheses.

$x (x - 5) (x + 6) = 0$

Step 6.2.2

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

$x = 0$

$x - 5 = 0$

$x + 6 = 0$

Step 6.2.3

Set $x$ equal to $0$ .

$x = 0$

Step 6.2.4

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 6.2.4.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 6.2.4.2

Add $5$ to both sides of the equation.

$x = 5$

Step 6.2.5

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 6.2.5.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 6.2.5.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 6.2.6

The final solution is all the values that make $x (x - 5) (x + 6) = 0$ true.

$x = 0, 5, - 6$

Step 6.3

Exclude the solutions that do not make $\frac{x^{3} + x^{2} - 30 x}{x^{2} - 36} = 0$ true.

$x = 0, 5$

Step 7

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

Interval Notation:

$(- \infty, - 6) \cup (- 6, - 3) \cup (- 3, 0) \cup (0, 5) \cup (5, 6) \cup (6, \infty)$

Set-Builder Notation:

${x | x \neq 0, - 3, 5, 6, - 6}$

Step 8