What is the sum of `(4x)/(x-1) and (4x+4)/(x^(2)-1)` expressed in simplest form?

Options

A. `(4x+1)/(x-1)`
B. `(4(x+1))/(x-1)`
C. `(4(x^(2)+4x-1))/(x^(2)-1)`
D. `(4(x+2))/(x^(2)-1)`

Answer

To find the sum of the two expressions \(\frac{4x}{x-1}\) and \(\frac{4x+4}{x^2-1}\), we will follow these steps:

### Step 1: Identify the expressions
We have:
\[ \frac{4x}{x-1} + \frac{4x+4}{x^2-1} \]

### Step 2: Factor the denominator of the second term
The denominator \(x^2 - 1\) can be factored using the difference of squares:
\[ x^2 - 1 = (x-1)(x+1) \]
Thus, we rewrite the second term:
\[ \frac{4x+4}{x^2-1} = \frac{4x+4}{(x-1)(x+1)} \]

### Step 3: Rewrite the first term with a common denominator
The first term \(\frac{4x}{x-1}\) can be rewritten to have the same denominator as the second term:
\[ \frac{4x}{x-1} = \frac{4x(x+1)}{(x-1)(x+1)} \]

### Step 4: Add the two fractions
Now we can add both fractions:
\[ \frac{4x(x+1)}{(x-1)(x+1)} + \frac{4x+4}{(x-1)(x+1)} \]
Combining the numerators:
\[ \frac{4x(x+1) + (4x+4)}{(x-1)(x+1)} \]

### Step 5: Simplify the numerator
Now we simplify the numerator:
\[ 4x(x+1) + (4x + 4) = 4x^2 + 4x + 4x + 4 = 4x^2 + 8x + 4 \]

### Step 6: Factor the numerator
We can factor out a 4 from the numerator:
\[ 4(x^2 + 2x + 1) = 4(x+1)^2 \]

### Step 7: Write the final expression
Now we can write the entire expression as:
\[ \frac{4(x+1)^2}{(x-1)(x+1)} \]

### Step 8: Cancel common factors
We can cancel the \((x+1)\) from the numerator and denominator (assuming \(x \neq -1\)):
\[ \frac{4(x+1)}{x-1} \]

### Final Answer
Thus, the sum of the two expressions in simplest form is:
\[ \frac{4(x+1)}{x-1} \]

---