Algebra Examples

$\frac{x + 3}{x^{2} + 8 x + 15} \div \frac{x^{2} - 25}{x - 5}$

Step 1

Set the denominator in $\frac{x + 3}{x^{2} + 8 x + 15}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 8 x + 15 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x^{2} + 8 x + 15$ 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 $8$ .

$3, 5$

Step 2.1.2

Write the factored form using these integers.

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

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 $0$ .

$x + 3 = 0$

$x + 5 = 0$

Step 2.3

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

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Step 2.3.1

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

$x + 3 = 0$

Step 2.3.2

Subtract $3$ from both sides of the equation.

$x = - 3$

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

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.5

The final solution is all the values that make $(x + 3) (x + 5) = 0$ true.

$x = - 3, - 5$

Step 3

Set the denominator in $\frac{x^{2} - 25}{x - 5}$ equal to $0$ to find where the expression is undefined.

$x - 5 = 0$

Step 4

Add $5$ to both sides of the equation.

$x = 5$

Step 5

Set the denominator in $\frac{x + 3}{x^{2} + 8 x + 15} \div \frac{x^{2} - 25}{x - 5}$ equal to $0$ to find where the expression is undefined.

$\frac{x^{2} - 25}{x - 5} = 0$

Step 6

Solve for $x$ .

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Step 6.1

Set the numerator equal to zero.

$x^{2} - 25 = 0$

Step 6.2

Solve the equation for $x$ .

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Step 6.2.1

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 6.2.2

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

$x = \pm \sqrt{25}$

Step 6.2.3

Simplify $\pm \sqrt{25}$ .

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Step 6.2.3.1

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

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

Step 6.2.3.2

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

$x = \pm 5$

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.

$x = 5$

Step 6.2.4.2

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

$x = - 5$

Step 6.2.4.3

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

$x = 5, - 5$

Step 6.3

Exclude the solutions that do not make $\frac{x^{2} - 25}{x - 5} = 0$ true.

$x = - 5$

Step 7

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 5, x = - 3, x = 5$

Step 8