Factor completely. \[ x^{3}+6 x^{2}+16 x+96 \]

To factor the polynomial \( x^{3}+6 x^{2}+16 x+96 \) completely, we can start by using the Rational Root Theorem to test for possible rational roots. Testing \( x = -2 \): \[ (-2)^{3} + 6(-2)^{2} + 16(-2) + 96 = -8 + 24 - 32 + 96 = 80 \quad \text{(not a root)} \] Next, testing \( x = -4 \): \[ (-4)^{3} + 6(-4)^{2} + 16(-4) + 96 = -64 + 96 - 64 + 96 = 64 \quad \text{(not a root)} \] Testing \( x = -6 \): \[ (-6)^{3} + 6(-6)^{2} + 16(-6) + 96 = -216 + 216 - 96 + 96 = 0 \quad \text{(this is a root)} \] Now that we have found \( x = -6 \) is a root, we can perform synthetic division with \( x + 6 \): \[ \begin{array}{r|rrrr} -6 & 1 & 6 & 16 & 96 \\ & & -6 & 0 & -96 \\ \hline & 1 & 0 & 16 & 0 \\ \end{array} \] This gives us \( x^2 + 16 \) as the quotient. Therefore, we have: \[ x^{3}+6 x^{2}+16 x+96 = (x + 6)(x^2 + 16) \] The term \( x^2 + 16 \) cannot be factored further over the real numbers, but it can be expressed with complex numbers: \[ x^2 + 16 = (x + 4i)(x - 4i) \] Thus, the completely factored form of the polynomial \( x^{3}+6 x^{2}+16 x+96 \) is: \[ (x + 6)(x + 4i)(x - 4i) \]