Linear Algebra Examples

$\frac{x^{2} - 25}{x^{2} - 2 x - 15} \cdot \frac{x + 3}{x}$

Step 1

Set the denominator in $\frac{x^{2} - 25}{x^{2} - 2 x - 15}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 2 x - 15 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x^{2} - 2 x - 15$ 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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Step 2.1.2

Write the factored form using these integers.

$(x - 5) (x + 3) = 0$

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 $0$ .

$x - 5 = 0$

$x + 3 = 0$

Step 2.3

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

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Step 2.3.1

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

$x - 5 = 0$

Step 2.3.2

Add $5$ to both sides of the equation.

$x = 5$

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 $(x - 5) (x + 3) = 0$ true.

$x = 5, - 3$

Step 3

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

$x = 0$

Step 4

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

Interval Notation:

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

Set-Builder Notation:

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

Step 5