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

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