Solve by Factoring 64/(x^2-16)+1=(2x)/(x-4)

$\frac{64}{x^{2} - 16} + 1 = \frac{2 x}{x - 4}$

Step 1

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

$\frac{64}{x^{2} - 16} + 1 - \frac{2 x}{x - 4} = 0$

Step 2

Simplify $\frac{64}{x^{2} - 16} + 1 - \frac{2 x}{x - 4}$ .

Simplify the denominator.

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

$\frac{64}{x^{2} - 4^{2}} + 1 - \frac{2 x}{x - 4} = 0$

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 $b = 4$ .

$\frac{64}{(x + 4) (x - 4)} + 1 - \frac{2 x}{x - 4} = 0$

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

$\frac{64}{(x + 4) (x - 4)} - \frac{2 x}{x - 4} \cdot \frac{x + 4}{x + 4} + 1 = 0$

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

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

$\frac{64}{(x + 4) (x - 4)} - \frac{2 x (x + 4)}{(x - 4) (x + 4)} + 1 = 0$

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

$\frac{64}{(x + 4) (x - 4)} - \frac{2 x (x + 4)}{(x + 4) (x - 4)} + 1 = 0$

Combine the numerators over the common denominator.

$\frac{64 - 2 x (x + 4)}{(x + 4) (x - 4)} + 1 = 0$

Simplify each term.

Simplify the numerator.

Factor $2$ out of $64 - 2 x (x + 4)$ .

Factor $2$ out of $64$ .

$\frac{2 (32) - 2 x (x + 4)}{(x + 4) (x - 4)} + 1 = 0$

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

$\frac{2 (32) + 2 (- x (x + 4))}{(x + 4) (x - 4)} + 1 = 0$

Factor $2$ out of $2 (32) + 2 (- x (x + 4))$ .

$\frac{2 (32 - x (x + 4))}{(x + 4) (x - 4)} + 1 = 0$

Apply the distributive property.

$\frac{2 (32 - x \cdot x - x \cdot 4)}{(x + 4) (x - 4)} + 1 = 0$

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

Move $x$ .

$\frac{2 (32 - (x \cdot x) - x \cdot 4)}{(x + 4) (x - 4)} + 1 = 0$

Multiply $x$ by $x$ .

$\frac{2 (32 - x^{2} - x \cdot 4)}{(x + 4) (x - 4)} + 1 = 0$

Multiply $4$ by $- 1$ .

$\frac{2 (32 - x^{2} - 4 x)}{(x + 4) (x - 4)} + 1 = 0$

Reorder terms.

$\frac{2 (- x^{2} - 4 x + 32)}{(x + 4) (x - 4)} + 1 = 0$

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 $a \cdot c = - 1 \cdot 32 = - 32$ and whose sum is $b = - 4$ .

Factor $- 4$ out of $- 4 x$ .

$\frac{2 (- x^{2} - 4 (x) + 32)}{(x + 4) (x - 4)} + 1 = 0$

Rewrite $- 4$ as $4$ plus $- 8$

$\frac{2 (- x^{2} + (4 - 8) x + 32)}{(x + 4) (x - 4)} + 1 = 0$

Apply the distributive property.

$\frac{2 (- x^{2} + 4 x - 8 x + 32)}{(x + 4) (x - 4)} + 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{2 ((- x^{2} + 4 x) - 8 x + 32)}{(x + 4) (x - 4)} + 1 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $- x + 4$ .

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

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

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

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

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

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

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 2$ by $8$ .

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

Add $- 2 x$ and $x$ .

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

Add $- 16$ and $4$ .

$\frac{- x - 12}{x + 4} = 0$

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

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

Rewrite $- 12$ as $- 1 (12)$ .

$\frac{- (x) - 1 (12)}{x + 4} = 0$

Factor $- 1$ out of $- (x) - 1 (12)$ .

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

Rewrite $- (x + 12)$ as $- 1 (x + 12)$ .

$\frac{- 1 (x + 12)}{x + 4} = 0$

Move the negative in front of the fraction.

$- \frac{x + 12}{x + 4} = 0$

Step 3

Set the numerator equal to zero.

$x + 12 = 0$

Step 4

Subtract $12$ from both sides of the equation.

$x = - 12$

Solve by Factoring 64/(x^2-16)+1=(2x)/(x-4)

Related Questions

Support LCMGCF.com by disabling your adblocker.