Solve by Factoring 1/(x+4)+x/(x-4)=-8/(x^2-16)

$\frac{1}{x + 4} + \frac{x}{x - 4} = \frac{- 8}{x^{2} - 16}$

Step 1

Subtract $\frac{- 8}{x^{2} - 16}$ from both sides of the equation.

$\frac{1}{x + 4} + \frac{x}{x - 4} - \frac{- 8}{x^{2} - 16} = 0$

Step 2

Simplify $\frac{1}{x + 4} + \frac{x}{x - 4} - \frac{- 8}{x^{2} - 16}$ .

Simplify each term.

Simplify the denominator.

Rewrite $16$ as $4^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$\frac{1}{x + 4} + \frac{x}{x - 4} - \frac{- 8}{(x + 4) (x - 4)} = 0$

Move the negative in front of the fraction.

$\frac{1}{x + 4} + \frac{x}{x - 4} - - \frac{8}{(x + 4) (x - 4)} = 0$

Multiply $- - \frac{8}{(x + 4) (x - 4)}$ .

Multiply $- 1$ by $- 1$ .

$\frac{1}{x + 4} + \frac{x}{x - 4} + 1 \frac{8}{(x + 4) (x - 4)} = 0$

Multiply $\frac{8}{(x + 4) (x - 4)}$ by $1$ .

$\frac{1}{x + 4} + \frac{x}{x - 4} + \frac{8}{(x + 4) (x - 4)} = 0$

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

$\frac{1}{x + 4} \cdot \frac{x - 4}{x - 4} + \frac{x}{x - 4} + \frac{8}{(x + 4) (x - 4)} = 0$

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

$\frac{1}{x + 4} \cdot \frac{x - 4}{x - 4} + \frac{x}{x - 4} \cdot \frac{x + 4}{x + 4} + \frac{8}{(x + 4) (x - 4)} = 0$

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $x$ by $x$ .

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

Move $4$ to the left of $x$ .

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

Add $x$ and $4 x$ .

$\frac{5 x - 4 + x^{2} + 8}{(x + 4) (x - 4)} = 0$

Add $- 4$ and $8$ .

$\frac{5 x + x^{2} + 4}{(x + 4) (x - 4)} = 0$

Simplify the numerator.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$\frac{5 u + u^{2} + 4}{(x + 4) (x - 4)} = 0$

Factor $5 u + u^{2} + 4$ 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 $4$ and whose sum is $5$ .

$1, 4$

Write the factored form using these integers.

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

Replace all occurrences of $u$ with $x$ .

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

Cancel the common factor of $x + 4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

Set the numerator equal to zero.

$x + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Solve by Factoring 1/(x+4)+x/(x-4)=-8/(x^2-16)

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