Precalculus Examples

$\frac{x^{2} + 5 x + 4}{x^{2} - 13 x + 40} \cdot \frac{x - 8}{x^{2} - 1}$

Step 1

Set the denominator in $\frac{x^{2} + 5 x + 4}{x^{2} - 13 x + 40}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 13 x + 40 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x^{2} - 13 x + 40$ 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 $40$ and whose sum is $- 13$ .

$- 8, - 5$

Step 2.1.2

Write the factored form using these integers.

$(x - 8) (x - 5) = 0$

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 $0$ .

$x - 8 = 0$

$x - 5 = 0$

Step 2.3

Set $x - 8$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Step 2.3.2

Add $8$ to both sides of the equation.

$x = 8$

Step 2.4

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

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Step 2.4.1

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

$x - 5 = 0$

Step 2.4.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.5

The final solution is all the values that make $(x - 8) (x - 5) = 0$ true.

$x = 8, 5$

Step 3

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

$x^{2} - 1 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 4.2

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

$x = \pm \sqrt{1}$

Step 4.3

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

Step 4.4.2

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

$x = - 1$

Step 4.4.3

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

$x = 1, - 1$

Step 5

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, 1) \cup (1, 5) \cup (5, 8) \cup (8, \infty)$

Set-Builder Notation:

${x | x \neq 1, - 1, 5, 8}$

Step 6