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

To expand the function \( f(x) = (x-3)^{4} \), 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, taking \( a = x \), \( b = -3 \), and \( n = 4 \), we get: \[ f(x) = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} (-3)^k \] Calculating each term: - For \( k = 0 \): \( \binom{4}{0} x^{4} (-3)^0 = 1 \cdot x^{4} \cdot 1 = x^{4} \) - For \( k = 1 \): \( \binom{4}{1} x^{3} (-3)^{1} = 4 \cdot x^{3} \cdot (-3) = -12x^{3} \) - For \( k = 2 \): \( \binom{4}{2} x^{2} (-3)^{2} = 6 \cdot x^{2} \cdot 9 = 54x^{2} \) - For \( k = 3 \): \( \binom{4}{3} x^{1} (-3)^{3} = 4 \cdot x \cdot (-27) = -108x \) - For \( k = 4 \): \( \binom{4}{4} x^{0} (-3)^{4} = 1 \cdot 1 \cdot 81 = 81 \) Combining all these terms, we get: \[ f(x) = x^{4} - 12x^{3} + 54x^{2} - 108x + 81 \] So, your expanded function is: \[ f(x) = x^{4} - 12x^{3} + 54x^{2} - 108x + 81 \]