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

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

Step 1

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

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

Step 2

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

Simplify the denominator.

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

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

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 $b = 2$ .

$\frac{16}{(x + 2) (x - 2)} + 1 - \frac{2 x}{x - 2} = 0$

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

$\frac{16}{(x + 2) (x - 2)} - \frac{2 x}{x - 2} \cdot \frac{x + 2}{x + 2} + 1 = 0$

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

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

$\frac{16}{(x + 2) (x - 2)} - \frac{2 x (x + 2)}{(x - 2) (x + 2)} + 1 = 0$

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

$\frac{16}{(x + 2) (x - 2)} - \frac{2 x (x + 2)}{(x + 2) (x - 2)} + 1 = 0$

Combine the numerators over the common denominator.

$\frac{16 - 2 x (x + 2)}{(x + 2) (x - 2)} + 1 = 0$

Simplify each term.

Simplify the numerator.

Factor $2$ out of $16 - 2 x (x + 2)$ .

Factor $2$ out of $16$ .

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

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

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

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

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

Apply the distributive property.

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

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $2$ by $- 1$ .

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

Reorder terms.

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

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 8 = - 8$ and whose sum is $b = - 2$ .

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

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

Rewrite $- 2$ as $2$ plus $- 4$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- x + 2$ .

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

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

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

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

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

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

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 2$ by $4$ .

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

Add $- 2 x$ and $x$ .

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

Add $- 8$ and $2$ .

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

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

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

Rewrite $- 6$ as $- 1 (6)$ .

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

Factor $- 1$ out of $- (x) - 1 (6)$ .

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

Rewrite $- (x + 6)$ as $- 1 (x + 6)$ .

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

Move the negative in front of the fraction.

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

Step 3

Set the numerator equal to zero.

$x + 6 = 0$

Step 4

Subtract $6$ from both sides of the equation.

$x = - 6$

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

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