Find all zeros of the function \( f(x)=4 x^{3}+32 x^{2}+76 x+48 \). Enter the zeros separated by commas.

To find the zeros of the function \( f(x) = 4x^3 + 32x^2 + 76x + 48 \), we can start by factoring out the greatest common factor, which is 4: \[ f(x) = 4(x^3 + 8x^2 + 19x + 12). \] Now, we need to find the roots of the cubic polynomial \( x^3 + 8x^2 + 19x + 12 \). We can attempt the Rational Root Theorem to find potential rational roots, which suggest trying \( x = -1, -2, -3, -4, -6, -12 \). Trying \( x = -3 \): \[ f(-3) = (-3)^3 + 8(-3)^2 + 19(-3) + 12 = -27 + 72 - 57 + 12 = 0. \] Thus, \( x = -3 \) is a root. Next, we can use synthetic division to divide \( x^3 + 8x^2 + 19x + 12 \) by \( x + 3 \): ``` -3 | 1 8 19 12 | -3 -15 -12 ------------------- 1 5 4 0 ``` This gives us \( x^2 + 5x + 4 \). Now we can factor \( x^2 + 5x + 4 \): \[ x^2 + 5x + 4 = (x + 1)(x + 4). \] Now we have the complete factorization: \[ f(x) = 4(x + 3)(x + 1)(x + 4). \] The zeros of the function are: \[ x + 3 = 0 \implies x = -3, \] \[ x + 1 = 0 \implies x = -1, \] \[ x + 4 = 0 \implies x = -4. \] Therefore, the function \( f(x) \) has zeros at: \[ \boxed{-3, -1, -4}. \]