Solve by Factoring z/(z-2)+1/(z+2)=8/(z^2-4)

$\frac{z}{z - 2} + \frac{1}{z + 2} = \frac{8}{z^{2} - 4}$

Step 1

Subtract $\frac{8}{z^{2} - 4}$ from both sides of the equation.

$\frac{z}{z - 2} + \frac{1}{z + 2} - \frac{8}{z^{2} - 4} = 0$

Step 2

Simplify $\frac{z}{z - 2} + \frac{1}{z + 2} - \frac{8}{z^{2} - 4}$ .

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$\frac{z}{z - 2} + \frac{1}{z + 2} - \frac{8}{z^{2} - 2^{2}} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = z$ and $b = 2$ .

$\frac{z}{z - 2} + \frac{1}{z + 2} - \frac{8}{(z + 2) (z - 2)} = 0$

To write $\frac{z}{z - 2}$ as a fraction with a common denominator, multiply by $\frac{z + 2}{z + 2}$ .

$\frac{z}{z - 2} \cdot \frac{z + 2}{z + 2} + \frac{1}{z + 2} - \frac{8}{(z + 2) (z - 2)} = 0$

To write $\frac{1}{z + 2}$ as a fraction with a common denominator, multiply by $\frac{z - 2}{z - 2}$ .

$\frac{z}{z - 2} \cdot \frac{z + 2}{z + 2} + \frac{1}{z + 2} \cdot \frac{z - 2}{z - 2} - \frac{8}{(z + 2) (z - 2)} = 0$

Write each expression with a common denominator of $(z - 2) (z + 2)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{z}{z - 2}$ by $\frac{z + 2}{z + 2}$ .

$\frac{z (z + 2)}{(z - 2) (z + 2)} + \frac{1}{z + 2} \cdot \frac{z - 2}{z - 2} - \frac{8}{(z + 2) (z - 2)} = 0$

Multiply $\frac{1}{z + 2}$ by $\frac{z - 2}{z - 2}$ .

$\frac{z (z + 2)}{(z - 2) (z + 2)} + \frac{z - 2}{(z + 2) (z - 2)} - \frac{8}{(z + 2) (z - 2)} = 0$

Reorder the factors of $(z - 2) (z + 2)$ .

$\frac{z (z + 2)}{(z + 2) (z - 2)} + \frac{z - 2}{(z + 2) (z - 2)} - \frac{8}{(z + 2) (z - 2)} = 0$

Combine the numerators over the common denominator.

$\frac{z (z + 2) + z - 2}{(z + 2) (z - 2)} - \frac{8}{(z + 2) (z - 2)} = 0$

Combine the numerators over the common denominator.

$\frac{z (z + 2) + z - 2 - 8}{(z + 2) (z - 2)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{z \cdot z + z \cdot 2 + z - 2 - 8}{(z + 2) (z - 2)} = 0$

Multiply $z$ by $z$ .

$\frac{z^{2} + z \cdot 2 + z - 2 - 8}{(z + 2) (z - 2)} = 0$

Move $2$ to the left of $z$ .

$\frac{z^{2} + 2 z + z - 2 - 8}{(z + 2) (z - 2)} = 0$

Add $2 z$ and $z$ .

$\frac{z^{2} + 3 z - 2 - 8}{(z + 2) (z - 2)} = 0$

Subtract $8$ from $- 2$ .

$\frac{z^{2} + 3 z - 10}{(z + 2) (z - 2)} = 0$

Factor $z^{2} + 3 z - 10$ using the AC method.

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 $- 10$ and whose sum is $3$ .

$- 2, 5$

Write the factored form using these integers.

$\frac{(z - 2) (z + 5)}{(z + 2) (z - 2)} = 0$

Cancel the common factor of $z - 2$ .

Cancel the common factor.

$\frac{(z - 2) (z + 5)}{(z + 2) (z - 2)} = 0$

Rewrite the expression.

$\frac{z + 5}{z + 2} = 0$

Step 3

Set the numerator equal to zero.

$z + 5 = 0$

Step 4

Subtract $5$ from both sides of the equation.

$z = - 5$

Solve by Factoring z/(z-2)+1/(z+2)=8/(z^2-4)

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