Expand the function. \( f(x)=(3 x+2)^{4} \) \( [?] x^{4}+\square x^{3}+\square x^{2}+\square x+\square \)

To expand the function \( f(x) = (3x + 2)^{4} \), we can use the binomial theorem, which states that \( (a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \). Here, \( a = 3x \), \( b = 2 \), and \( n = 4 \). Expanding this gives: \[ f(x) = \binom{4}{0} (3x)^4 (2)^0 + \binom{4}{1} (3x)^3 (2)^1 + \binom{4}{2} (3x)^2 (2)^2 + \binom{4}{3} (3x)^1 (2)^3 + \binom{4}{4} (3x)^0 (2)^4 \] Calculating each term gives: 1. \( 1 \cdot (3x)^4 \cdot 1 = 81x^4 \) 2. \( 4 \cdot (3x)^3 \cdot 2 = 216x^3 \) 3. \( 6 \cdot (3x)^2 \cdot 4 = 216x^2 \) 4. \( 4 \cdot (3x)^1 \cdot 8 = 96x \) 5. \( 1 \cdot 1 \cdot 16 = 16 \) Putting it all together, we have: \[ f(x) = 81x^{4} + 216x^{3} + 216x^{2} + 96x + 16 \] So, the expanded function is: \[ f(x) = 81x^4 + 216x^3 + 216x^2 + 96x + 16 \]