Find the product. \( \frac{5 x^{2}-40 x}{x^{2}-4 x-32} \cdot \frac{x^{2}-x-20}{5 x}=\square, x \neq \square, x \neq \square, x \neq \square \)

To find the product \[ \frac{5 x^{2}-40 x}{x^{2}-4 x-32} \cdot \frac{x^{2}-x-20}{5 x}, \] we start by factoring each polynomial. 1. **Factoring the numerators and denominators:** For the first numerator, \(5x^2 - 40x\): \[ 5x(x - 8). \] For the first denominator, \(x^2 - 4x - 32\): This can be factored by finding two numbers that multiply to \(-32\) and add to \(-4\), which are \(-8\) and \(4\): \[ (x - 8)(x + 4). \] For the second numerator, \(x^2 - x - 20\): The numbers \(-5\) and \(4\) work, so we have: \[ (x - 5)(x + 4). \] The second denominator is simply \(5x\). 2. **Putting it all together:** Now substituting all the factored forms: \[ \frac{5x(x - 8)}{(x - 8)(x + 4)} \cdot \frac{(x - 5)(x + 4)}{5x}. \] 3. **Canceling common factors:** We can cancel out \(5x\) in the numerator and denominator, as well as \(x - 8\) and \(x + 4\): After canceling, we are left with: \[ \frac{x - 5}{1} = x - 5. \] 4. **Determining restrictions**: We need to figure out the restrictions based on our factored expressions, they correspond to values that will make the original denominators equal to zero: - From \(x^2 - 4x - 32 = 0\), we find \(x = 8\) and \(x = -4\). - From \(5x = 0\), we find \(x = 0\). Thus, the final statement is \[ \frac{5 x^{2}-40 x}{x^{2}-4 x-32} \cdot \frac{x^{2}-x-20}{5 x} = x - 5, x \neq 8, x \neq -4, x \neq 0. \]