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

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 \( n = 4 \). Applying the theorem: \[ f(x) = \sum_{k=0}^{4} \binom{4}{k} (3x)^{4-k} (2)^{k} \] Calculating each term: - For \( k = 0 \): \( \binom{4}{0} (3x)^{4} (2)^{0} = 81x^4 \) - For \( k = 1 \): \( \binom{4}{1} (3x)^{3} (2)^{1} = 4 \cdot 27x^3 \cdot 2 = 216x^3 \) - For \( k = 2 \): \( \binom{4}{2} (3x)^{2} (2)^{2} = 6 \cdot 9x^2 \cdot 4 = 216x^2 \) - For \( k = 3 \): \( \binom{4}{3} (3x)^{1} (2)^{3} = 4 \cdot 3x \cdot 8 = 96x \) - For \( k = 4 \): \( \binom{4}{4} (3x)^{0} (2)^{4} = 16 \) Putting it all together: \[ f(x) = 81x^4 + 216x^3 + 216x^2 + 96x + 16 \] So the expanded form is: \[ f(x) = 81x^4 + 216x^3 + 216x^2 + 96x + 16 \]