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

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

Move $\frac{2 x}{x - 2}$ to the left side of the equation by subtracting it from both sides.

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

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

Tap for more steps…

Simplify the denominator.

Tap for more steps…

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 $1$ .

Tap for more steps…

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

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

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.

Tap for more steps…

Simplify the numerator.

Tap for more steps…

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

Tap for more steps…

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.

Tap for more steps…

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.

Tap for more steps…

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 $b = - 2$ .

Tap for more steps…

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.

Tap for more steps…

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 $x - 2$ .

Tap for more steps…

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.

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

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.

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

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

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

Simplify terms.

Tap for more steps…

Combine $- 1$ and $\frac{x + 2}{x + 2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Tap for more steps…

Factor $- 1$ out of $- 2 (x + 4) - (x + 2)$ .

Tap for more steps…

Factor $- 1$ out of $- 2 (x + 4)$ .

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

Factor $- 1$ out of $- (2 (x + 4)) - (x + 2)$ .

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

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 $x$ .

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

Add $8$ and $2$ .

$\frac{- (3 x + 10)}{x + 2} = 0$

Move the negative in front of the fraction.

$- \frac{3 x + 10}{x + 2} = 0$

Simplify the denominator.

Tap for more steps…

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 $1$ .

Tap for more steps…

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

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

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.

Tap for more steps…

Simplify the numerator.

Tap for more steps…

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

Tap for more steps…

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.

Tap for more steps…

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.

Tap for more steps…

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 $b = - 2$ .

Tap for more steps…

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.

Tap for more steps…

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 $x - 2$ .

Tap for more steps…

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

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

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

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

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.

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

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.

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

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

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

Simplify terms.

Tap for more steps…

Combine $- 1$ and $\frac{x + 2}{x + 2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Tap for more steps…

Factor $- 1$ out of $- 2 (x + 4) - (x + 2)$ .

Tap for more steps…

Factor $- 1$ out of $- 2 (x + 4)$ .

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

Factor $- 1$ out of $- (2 (x + 4)) - (x + 2)$ .

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

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 $x$ .

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

Add $8$ and $2$ .

$\frac{- (3 x + 10)}{x + 2} = 0$

Move the negative in front of the fraction.

$- \frac{3 x + 10}{x + 2} = 0$

Find the LCD of the terms in the equation.

Tap for more steps…

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x + 2, 1$

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factor for $x + 2$ is $x + 2$ itself.

$(x + 2) = x + 2$

$(x + 2)$ occurs $1$ time.

The LCM of $x + 2$ is the result of multiplying all factors the greatest number of times they occur in either term.

$x + 2$

Multiply each term by $x + 2$ and simplify.

Tap for more steps…

Multiply each term in $- \frac{3 x + 10}{x + 2} = 0$ by $x + 2$ in order to remove all the denominators from the equation.

$- \frac{3 x + 10}{x + 2} \cdot (x + 2) = 0 \cdot (x + 2)$

Simplify $- \frac{3 x + 10}{x + 2} \cdot (x + 2)$ .

Tap for more steps…

Cancel the common factor of $x + 2$ .

Tap for more steps…

Move the leading negative in $- \frac{3 x + 10}{x + 2}$ into the numerator.

$\frac{- (3 x + 10)}{x + 2} \cdot (x + 2) = 0 \cdot (x + 2)$

Cancel the common factor.

$\frac{- (3 x + 10)}{x + 2} \cdot (x + 2) = 0 \cdot (x + 2)$

Rewrite the expression.

$- (3 x + 10) = 0 \cdot (x + 2)$

Apply the distributive property.

$- (3 x) - 1 \cdot 10 = 0 \cdot (x + 2)$

Multiply.

Tap for more steps…

Multiply $3$ by $- 1$ .

$- 3 x - 1 \cdot 10 = 0 \cdot (x + 2)$

Multiply $- 1$ by $10$ .

$- 3 x - 10 = 0 \cdot (x + 2)$

Multiply $0$ by $x + 2$ .

$- 3 x - 10 = 0$

Solve the equation.

Tap for more steps…

Add $10$ to both sides of the equation.

$- 3 x = 10$

Divide each term by $- 3$ and simplify.

Tap for more steps…

Divide each term in $- 3 x = 10$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{10}{- 3}$

Cancel the common factor of $- 3$ .

Tap for more steps…

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{10}{- 3}$

Divide $x$ by $1$ .

$x = \frac{10}{- 3}$

Move the negative in front of the fraction.

$x = - \frac{10}{3}$

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

Download our
App from the store

Create a High Performed UI/UX Design from a Silicon Valley.

Download our App from the store

Create a High Performed UI/UX Design from a Silicon Valley.

Download our
App from the store