Use the Binomial Theorem to expand \( (x+3)^{5} \) \( \begin{array}{l}x^{5}+3 x^{4}+9 x^{3}+27 x^{2}+81 x+243 \\ \frac{x^{5}+15 x^{4}+90 x^{3}+270 x^{2}+405 x+243}{x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+5 x+1} \\ x^{2}+243\end{array} \)

To expand \( (x+3)^{5} \) using the Binomial Theorem, we can use the formula: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] In this case, \( a = x \), \( b = 3 \), and \( n = 5 \). Thus, we have: \[ (x + 3)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} \cdot 3^k \] Now, we will calculate each term in the expansion: - For \( k = 0 \): \[ \binom{5}{0} x^{5-0} \cdot 3^0 = 1 \cdot x^5 \cdot 1 = x^5 \] - For \( k = 1 \): \[ \binom{5}{1} x^{5-1} \cdot 3^1 = 5 \cdot x^4 \cdot 3 = 15x^4 \] - For \( k = 2 \): \[ \binom{5}{2} x^{5-2} \cdot 3^2 = 10 \cdot x^3 \cdot 9 = 90x^3 \] - For \( k = 3 \): \[ \binom{5}{3} x^{5-3} \cdot 3^3 = 10 \cdot x^2 \cdot 27 = 270x^2 \] - For \( k = 4 \): \[ \binom{5}{4} x^{5-4} \cdot 3^4 = 5 \cdot x^1 \cdot 81 = 405x \] - For \( k = 5 \): \[ \binom{5}{5} x^{5-5} \cdot 3^5 = 1 \cdot 1 \cdot 243 = 243 \] Now, we can combine all these terms together: \[ (x + 3)^5 = x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243 \] Thus, the expansion of \( (x + 3)^5 \) is: \[ \boxed{x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243} \]