When dividing \( x^{3}+n x^{2}+4 n x-6 \) by \( x+3 \), the remainder is -48 . What is the value of \( n \) ? \( n=\square \) (Type an integer or a simplified fraction.)

To find the value of \( n \), we can use the Remainder Theorem, which states that the remainder of the division of a polynomial \( f(x) \) by \( x-c \) is \( f(c) \). In this problem, we are dividing \( x^{3}+n x^{2}+4 n x-6 \) by \( x+3 \), so we will evaluate the polynomial at \( x = -3 \). Let \( f(x) = x^{3} + n x^{2} + 4 n x - 6 \). According to the Remainder Theorem: \[ f(-3) = (-3)^{3} + n(-3)^{2} + 4n(-3) - 6 \] Calculating \( f(-3) \): \[ f(-3) = -27 + 9n - 12n - 6 = -27 - 3n - 6 = -33 - 3n \] We know that the remainder is -48: \[ -33 - 3n = -48 \] Now, we can solve for \( n \): \[ -3n = -48 + 33 \] \[ -3n = -15 \] \[ n = 5 \] Thus, the value of \( n \) is \( 5 \).