Multiply. \[ \frac{x^{2}-4 x+3}{x+1} \cdot \frac{x+3}{4 x-4} \] Simplify. your answer as much as possible.

To simplify the expression \[ \frac{x^{2}-4 x+3}{x+1} \cdot \frac{x+3}{4 x-4}, \] we start by factoring wherever possible. 1. Factor \(x^{2} - 4x + 3\): \[ x^{2} - 4x + 3 = (x-1)(x-3) \] 2. Observe that \(4x - 4\) can be factored as: \[ 4x - 4 = 4(x-1) \] Now we can rewrite the expression with these factors: \[ \frac{(x-1)(x-3)}{x+1} \cdot \frac{x+3}{4(x-1)} \] Next, we multiply the fractions: \[ \frac{(x-1)(x-3)(x+3)}{(x+1) \cdot 4(x-1)} \] Now, we can cancel \(x - 1\) in the numerator and denominator (assuming \(x \neq 1\)). This leaves us with: \[ \frac{(x-3)(x+3)}{4(x+1)} \] Next, we can simplify the numerator \( (x-3)(x+3) \) which becomes: \[ \frac{x^2 - 9}{4(x+1)} \] Thus, the simplified expression for our original multiplication is: \[ \frac{x^2 - 9}{4(x+1)} \] Therefore, the final simplified form is: \[ \frac{x^2 - 9}{4(x + 1)} \] This simplifies as far as possible, provided \(x \neq 1\) and \(x \neq -1\).