Find the Excluded Values 1/(x-4)-3/(x+2)=6/(x^2-2x-8)

$\frac{1}{x - 4} - \frac{3}{x + 2} = \frac{6}{x^{2} - 2 x - 8}$

Step 1

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

$\frac{1}{x - 4} - \frac{3}{x + 2} - \frac{6}{x^{2} - 2 x - 8} = 0$

Step 2

Simplify $\frac{1}{x - 4} - \frac{3}{x + 2} - \frac{6}{x^{2} - 2 x - 8}$ .

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 $- 2$ .

$- 4, 2$

Write the factored form using these integers.

$\frac{1}{x - 4} - \frac{3}{x + 2} - \frac{6}{(x - 4) (x + 2)} = 0$

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

$\frac{1}{x - 4} \cdot \frac{x + 2}{x + 2} - \frac{3}{x + 2} - \frac{6}{(x - 4) (x + 2)} = 0$

To write $- \frac{3}{x + 2}$ as a fraction with a common denominator, multiply by $\frac{x - 4}{x - 4}$ .

$\frac{1}{x - 4} \cdot \frac{x + 2}{x + 2} - \frac{3}{x + 2} \cdot \frac{x - 4}{x - 4} - \frac{6}{(x - 4) (x + 2)} = 0$

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

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

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

Multiply $\frac{3}{x + 2}$ by $\frac{x - 4}{x - 4}$ .

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

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

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

Combine the numerators over the common denominator.

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 3$ by $- 4$ .

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

Subtract $3 x$ from $x$ .

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

Add $2$ and $12$ .

$\frac{- 2 x + 14 - 6}{(x + 2) (x - 4)} = 0$

Subtract $6$ from $14$ .

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

Factor $2$ out of $- 2 x + 8$ .

Factor $2$ out of $- 2 x$ .

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

Factor $2$ out of $8$ .

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

Factor $2$ out of $2 (- x) + 2 (4)$ .

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

Cancel the common factor of $- x + 4$ and $x - 4$ .

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

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

Rewrite $4$ as $- 1 (- 4)$ .

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

Factor $- 1$ out of $- (x) - 1 (- 4)$ .

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

Rewrite $- (x - 4)$ as $- 1 (x - 4)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $2$ by $- 1$ .

$\frac{- 2}{x + 2} = 0$

Move the negative in front of the fraction.

$- \frac{2}{x + 2} = 0$

Step 3

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

$x + 2 = 0$

Step 4

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

Find the Excluded Values 1/(x-4)-3/(x+2)=6/(x^2-2x-8)

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