Precalculus Examples

$\frac{b x + 3 b}{x^{2} - 25} \cdot \frac{{(x - 5)}^{2}}{a x + 3 a}$

Step 1

Set the denominator in $\frac{b x + 3 b}{x^{2} - 25}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 25 = 0$

Step 2

Solve for $b$ .

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Step 2.1

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 2.2

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

$x = \pm \sqrt{25}$

Step 2.3

Simplify $\pm \sqrt{25}$ .

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Step 2.3.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 2.3.2

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

$x = \pm 5$

Step 2.4

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

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Step 2.4.1

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

$x = 5$

Step 2.4.2

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

$x = - 5$

Step 2.4.3

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

$x = 5, - 5$

Step 3

Set the denominator in $\frac{{(x - 5)}^{2}}{a x + 3 a}$ equal to $0$ to find where the expression is undefined.

$a x + 3 a = 0$

Step 4

Solve for $b$ .

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Step 4.1

Factor $a$ out of $a x + 3 a$ .

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Step 4.1.1

Factor $a$ out of $a x$ .

$a (x) + 3 a = 0$

Step 4.1.2

Factor $a$ out of $3 a$ .

$a (x) + a \cdot 3 = 0$

Step 4.1.3

Factor $a$ out of $a (x) + a \cdot 3$ .

$a (x + 3) = 0$

Step 4.2

Divide each term in $a (x + 3) = 0$ by $x + 3$ and simplify.

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Step 4.2.1

Divide each term in $a (x + 3) = 0$ by $x + 3$ .

$\frac{a (x + 3)}{x + 3} = \frac{0}{x + 3}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $x + 3$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{a (x + 3)}{x + 3} = \frac{0}{x + 3}$

Step 4.2.2.1.2

Divide $a$ by $1$ .

$a = \frac{0}{x + 3}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $0$ by $x + 3$ .

$a = 0$

Step 5

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

Interval Notation:

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

Set-Builder Notation:

${b | b \neq 0, 5, - 5}$