Find the Excluded Values 1/(x-2)-4/(x+4)=6/(x^2+2x-8)

$\frac{1}{x - 2} - \frac{4}{x + 4} = \frac{6}{x^{2} + 2 x - 8}$

Step 1

Subtract $\frac{6}{x^{2} + 2 x - 8}$ from both sides of the equation.

$\frac{1}{x - 2} - \frac{4}{x + 4} - \frac{6}{x^{2} + 2 x - 8} = 0$

Step 2

Simplify $\frac{1}{x - 2} - \frac{4}{x + 4} - \frac{6}{x^{2} + 2 x - 8}$ .

Factor $x^{2} + 2 x - 8$ 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 $- 8$ and whose sum is $2$ .

$- 2, 4$

Write the factored form using these integers.

$\frac{1}{x - 2} - \frac{4}{x + 4} - \frac{6}{(x - 2) (x + 4)} = 0$

To write $\frac{1}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{x + 4}{x + 4}$ .

$\frac{1}{x - 2} \cdot \frac{x + 4}{x + 4} - \frac{4}{x + 4} - \frac{6}{(x - 2) (x + 4)} = 0$

To write $- \frac{4}{x + 4}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$\frac{1}{x - 2} \cdot \frac{x + 4}{x + 4} - \frac{4}{x + 4} \cdot \frac{x - 2}{x - 2} - \frac{6}{(x - 2) (x + 4)} = 0$

Write each expression with a common denominator of $(x - 2) (x + 4)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{x - 2}$ by $\frac{x + 4}{x + 4}$ .

$\frac{x + 4}{(x - 2) (x + 4)} - \frac{4}{x + 4} \cdot \frac{x - 2}{x - 2} - \frac{6}{(x - 2) (x + 4)} = 0$

Multiply $\frac{4}{x + 4}$ by $\frac{x - 2}{x - 2}$ .

$\frac{x + 4}{(x - 2) (x + 4)} - \frac{4 (x - 2)}{(x + 4) (x - 2)} - \frac{6}{(x - 2) (x + 4)} = 0$

Reorder the factors of $(x - 2) (x + 4)$ .

$\frac{x + 4}{(x + 4) (x - 2)} - \frac{4 (x - 2)}{(x + 4) (x - 2)} - \frac{6}{(x - 2) (x + 4)} = 0$

Combine the numerators over the common denominator.

$\frac{x + 4 - 4 (x - 2)}{(x + 4) (x - 2)} - \frac{6}{(x - 2) (x + 4)} = 0$

Combine the numerators over the common denominator.

$\frac{x + 4 - 4 (x - 2) - 6}{(x + 4) (x - 2)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{x + 4 - 4 x - 4 \cdot - 2 - 6}{(x + 4) (x - 2)} = 0$

Multiply $- 4$ by $- 2$ .

$\frac{x + 4 - 4 x + 8 - 6}{(x + 4) (x - 2)} = 0$

Subtract $4 x$ from $x$ .

$\frac{- 3 x + 4 + 8 - 6}{(x + 4) (x - 2)} = 0$

Add $4$ and $8$ .

$\frac{- 3 x + 12 - 6}{(x + 4) (x - 2)} = 0$

Subtract $6$ from $12$ .

$\frac{- 3 x + 6}{(x + 4) (x - 2)} = 0$

Factor $3$ out of $- 3 x + 6$ .

Factor $3$ out of $- 3 x$ .

$\frac{3 (- x) + 6}{(x + 4) (x - 2)} = 0$

Factor $3$ out of $6$ .

$\frac{3 (- x) + 3 (2)}{(x + 4) (x - 2)} = 0$

Factor $3$ out of $3 (- x) + 3 (2)$ .

$\frac{3 (- x + 2)}{(x + 4) (x - 2)} = 0$

Cancel the common factor of $- x + 2$ and $x - 2$ .

Factor $- 1$ out of $- x$ .

$\frac{3 (- (x) + 2)}{(x + 4) (x - 2)} = 0$

Rewrite $2$ as $- 1 (- 2)$ .

$\frac{3 (- (x) - 1 (- 2))}{(x + 4) (x - 2)} = 0$

Factor $- 1$ out of $- (x) - 1 (- 2)$ .

$\frac{3 (- (x - 2))}{(x + 4) (x - 2)} = 0$

Rewrite $- (x - 2)$ as $- 1 (x - 2)$ .

$\frac{3 (- 1 (x - 2))}{(x + 4) (x - 2)} = 0$

Cancel the common factor.

$\frac{3 (- 1 (x - 2))}{(x + 4) (x - 2)} = 0$

Rewrite the expression.

$\frac{3 \cdot (- 1)}{x + 4} = 0$

Multiply $3$ by $- 1$ .

$\frac{- 3}{x + 4} = 0$

Move the negative in front of the fraction.

$- \frac{3}{x + 4} = 0$

Step 3

Set the denominator in $\frac{3}{x + 4}$ equal to $0$ to find where the expression is undefined.

$x + 4 = 0$

Step 4

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

Find the Excluded Values 1/(x-2)-4/(x+4)=6/(x^2+2x-8)

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