What is the product? \( \frac{x^{2}-16}{2 x+8} \cdot \frac{x^{3}-2 x^{2}+x}{x^{2}+3 x-4} \) \( \frac{x(x-4)(x-1)}{2(x+4)} \) \( \frac{x(x-1)}{2} \) \( \frac{(x+4)(x-4)}{2 x(x-1)} \) \( \frac{(x-4)(x-1)}{2 x(x+4)} \)

To simplify and determine the product of the expression \[ \frac{x^{2}-16}{2 x+8} \cdot \frac{x^{3}-2 x^{2}+x}{x^{2}+3 x-4}, \] we can start by factoring both numerators and denominators: 1. \( x^{2}-16 = (x-4)(x+4) \) (difference of squares). 2. \( 2x+8 = 2(x+4) \) (factoring out 2). 3. \( x^{3}-2x^{2}+x = x(x^{2}-2x+1) = x(x-1)^{2} \) (factoring out \(x\)). 4. \( x^{2}+3x-4 = (x+4)(x-1) \) (factoring the quadratic). Now substituting these factors into the original expression, we get: \[ \frac{(x-4)(x+4)}{2(x+4)} \cdot \frac{x(x-1)^{2}}{(x+4)(x-1)}. \] Next, we notice that we can cancel \( (x+4) \) from the numerator and denominator: \[ \frac{(x-4)}{2} \cdot \frac{x(x-1)}{(x-1)}. \] Then, we can cancel \( (x-1) \) as well from the second part: \[ \frac{(x-4)}{2} \cdot x. \] Thus, the product simplifies to: \[ \frac{x(x-4)}{2}. \] So, this is the simplest form of the product. Now to answer your request for additional knowledge... Did you know that factoring is not just a strategy for math problems but also a historical method that has roots in ancient civilizations? For instance, Babylonian mathematicians were using similar methods of simplification over 4000 years ago! They had impressive skills in algebra, and their clay tablets featured problems that required similar factoring techniques. For those eager to dive deeper into the world of algebra, there are some excellent resources! "Algebra Unplugged" by Kenn Amdahl is a fun and engaging read that demystifies algebra concepts with witty humor. Additionally, websites like Khan Academy offer free courses on algebra that can make even the most complex problems seem manageable and entertaining!