15) \( 6 x^{2}+96 x y+378 y \)

We start with the expression \[ 6x^2 + 96xy + 378y. \] **Step 1. Factor out the greatest common factor (GCF).** Notice that the coefficients \(6\), \(96\), and \(378\) are all divisible by \(6\). Factoring \(6\) out, we have \[ 6x^2 + 96xy + 378y = 6\left(x^2 + 16xy + 63y\right). \] **Step 2. Examine the expression inside the parentheses.** The expression inside is \[ x^2 + 16xy + 63y. \] Since the first term is \(x^2\) and the other terms both have a factor of \(y\), we can try to factor by grouping. Notice, however, that \(y\) is not present in \(x^2\) so a common factor \(y\) cannot be factored out from all three terms. Also, if we consider the expression as a quadratic in \(x\) with \(y\) acting as a fixed parameter, then the quadratic would be \[ x^2 + (16y)x + 63y. \] To factor a quadratic of the form \[ x^2 + bx + c, \] we look for two numbers that multiply to \(c\) and add to \(b\). Here, \(c = 63y\) and \(b = 16y\). If we try to write the two numbers as \(7y\) and \(9y\), then \[ 7y + 9y = 16y \quad \text{and} \quad (7y)(9y) = 63y^2. \] However, the product is \(63y^2\) rather than \(63y\), so the factorization is not possible in this way. **Step 3. Write the final factorization.** Since there is no further factorization possible using common factors or standard quadratic techniques, the completely factored form is \[ 6\left(x^2 + 16xy + 63y\right). \] Thus, the final answer is \[ \boxed{6\left(x^2 + 16xy + 63y\right)}. \]