Multiply. \[ \frac{2 x-8 y}{x^{2}-9 y^{2}} \cdot \frac{x^{2}-3 x y-18 y^{2}}{x-4 y} \] Simplify your answer as much as possible.

To simplify the expression \[ \frac{2 x-8 y}{x^{2}-9 y^{2}} \cdot \frac{x^{2}-3 x y-18 y^{2}}{x-4 y}, \] we can start by factoring the components in each fraction. 1. **Factoring the numerator and denominator**: - For the first term, \(2x - 8y\) can be factored as: \[ 2(x - 4y). \] - For the denominator \(x^2 - 9y^2\), this is a difference of squares: \[ x^2 - 9y^2 = (x - 3y)(x + 3y). \] - In the second fraction, \(x^2 - 3xy - 18y^2\) can be factored. To do this, we look for two numbers that multiply to \(-18\) and add to \(-3\). These numbers are \(-6\) and \(3\): \[ x^2 - 3xy - 18y^2 = (x - 6y)(x + 3y). \] - The denominator \(x - 4y\) is already in its simplest form, so we leave it as is. 2. **Putting it all together**: Now we can rewrite the entire expression with the factored terms: \[ \frac{2(x - 4y)}{(x - 3y)(x + 3y)} \cdot \frac{(x - 6y)(x + 3y)}{x - 4y}. \] 3. **Canceling common factors**: Notice that \(x - 4y\) in the numerator of the first fraction and \(x - 4y\) in the denominator of the second fraction cancel out: \[ = \frac{2 \cdot (x - 6y)(x + 3y)}{(x - 3y)(x + 3y)}. \] Next, we see that \(x + 3y\) cancels out from the numerator and the denominator: \[ = \frac{2(x - 6y)}{x - 3y}. \] 4. **Final result**: Thus, the simplified result is: \[ \frac{2(x - 6y)}{x - 3y}. \]