Solve by Factoring 2/(c-2)=2-4/c

$\frac{2}{c - 2} = 2 - \frac{4}{c}$

Step 1

Move all the expressions to the left side of the equation.

Subtract $2$ from both sides of the equation.

$\frac{2}{c - 2} - 2 = - \frac{4}{c}$

Add $\frac{4}{c}$ to both sides of the equation.

$\frac{2}{c - 2} - 2 + \frac{4}{c} = 0$

Step 2

Simplify $\frac{2}{c - 2} - 2 + \frac{4}{c}$ .

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

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

Combine $- 2$ and $\frac{c - 2}{c - 2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Factor $2$ out of $2 - 2 (c - 2)$ .

Factor $2$ out of $2$ .

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

Factor $2$ out of $- 2 (c - 2)$ .

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

Factor $2$ out of $2 (1) + 2 (- (c - 2))$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $- 2$ .

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

Add $1$ and $2$ .

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

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

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

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

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

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

Multiply $\frac{2 (- c + 3)}{c - 2}$ by $\frac{c}{c}$ .

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

Multiply $\frac{4}{c}$ by $\frac{c - 2}{c - 2}$ .

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

Reorder the factors of $(c - 2) c$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Factor $2$ out of $2 (- c + 3) c + 4 (c - 2)$ .

Factor $2$ out of $2 (- c + 3) c$ .

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

Factor $2$ out of $4 (c - 2)$ .

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

Factor $2$ out of $2 ((- c + 3) c) + 2 (2 (c - 2))$ .

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

Apply the distributive property.

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

Multiply $c$ by $c$ by adding the exponents.

Move $c$ .

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

Multiply $c$ by $c$ .

$\frac{2 (- c^{2} + 3 c + 2 (c - 2))}{c (c - 2)} = 0$

Apply the distributive property.

$\frac{2 (- c^{2} + 3 c + 2 c + 2 \cdot - 2)}{c (c - 2)} = 0$

Multiply $2$ by $- 2$ .

$\frac{2 (- c^{2} + 3 c + 2 c - 4)}{c (c - 2)} = 0$

Add $3 c$ and $2 c$ .

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

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 $a \cdot c = - 1 \cdot - 4 = 4$ and whose sum is $b = 5$ .

Factor $5$ out of $5 c$ .

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

Rewrite $5$ as $1$ plus $4$

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

Apply the distributive property.

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

Multiply $c$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $- c + 1$ .

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

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

Factor $- 1$ out of $- c$ .

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

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

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

Factor $- 1$ out of $- (c) - 1 (- 1)$ .

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

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

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

Move the negative in front of the fraction.

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

Reorder factors in $- \frac{(2 (c - 1)) (c - 4)}{c (c - 2)}$ .

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

Step 3

Set the numerator equal to zero.

$2 (c - 1) (c - 4) = 0$

Step 4

Solve the equation for $c$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$c - 1 = 0$

$c - 4 = 0$

Set $c - 1$ equal to $0$ and solve for $c$ .

Set $c - 1$ equal to $0$ .

$c - 1 = 0$

Add $1$ to both sides of the equation.

$c = 1$

Set $c - 4$ equal to $0$ and solve for $c$ .

Set $c - 4$ equal to $0$ .

$c - 4 = 0$

Add $4$ to both sides of the equation.

$c = 4$

The final solution is all the values that make $2 (c - 1) (c - 4) = 0$ true.

$c = 1, 4$

Solve by Factoring 2/(c-2)=2-4/c

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