Perform the following operation and express in simplest form. \[ \frac{x^{2}-1}{2 x+2} \div \frac{x^{2}-6 x+5}{2 x^{2}} \]

To solve the given expression \[ \frac{x^{2}-1}{2x+2} \div \frac{x^{2}-6x+5}{2x^{2}}, \] we start by rewriting the division as multiplication by the reciprocal: \[ \frac{x^{2}-1}{2x+2} \cdot \frac{2x^{2}}{x^{2}-6x+5}. \] Next, we factor the numerator and denominator of each fraction. 1. The first numerator \(x^{2} - 1\) is a difference of squares: \[ x^{2} - 1 = (x - 1)(x + 1). \] 2. The denominator \(2x + 2\) can be factored out: \[ 2x + 2 = 2(x + 1). \] 3. The second numerator \(2x^{2}\) remains as is: \[ 2x^{2} \text{ (no factoring needed)}. \] 4. The second denominator \(x^{2} - 6x + 5\) factors as: \[ x^{2} - 6x + 5 = (x - 1)(x - 5). \] Putting this all together, we rewrite the expression: \[ \frac{(x - 1)(x + 1)}{2(x + 1)} \cdot \frac{2x^{2}}{(x - 1)(x - 5)}. \] Now we can simplify by canceling common factors: - \(x + 1\) appears in both the numerator and the denominator. - \(2\) in the numerator and denominator cancels out. - \(x - 1\) appears in both the numerator and denominator. After cancelling, we have: \[ \frac{x^{2}}{x - 5}. \] Thus, the expression simplifies to: \[ \frac{x^{2}}{x - 5}. \] This is the simplest form. To summarize: \[ \boxed{\frac{x^{2}}{x - 5}} \]