To find the second term in the expansion of $(x + 3)^5$, we can use the binomial theorem. The binomial theorem states:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Here, $a = x$, $b = 3$, and $n = 5$. The general term in the expansion is given by:

$\text{Term}_{k+1} = \binom{n}{k} \cdot a^{n-k} \cdot b^k$

For the second term, $k = 1$:

$\text{2nd Term} = \binom{5}{1} \cdot x^{5-1} \cdot 3^1$

$= 5 \cdot x^4 \cdot 3 = 15x^4$

So, the second term in the expansion of $(x + 3)^5$ is $15x^4$.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:

1. What is the third term in the expansion of $(x + 3)^5$?
2. How would you find the fourth term in the expansion of $(2x + 1)^6$?
3. What is the coefficient of $x^3$ in the expansion of $(x + 4)^7$?
4. Can you derive the binomial expansion formula?
5. How would the expansion change if the exponent were negative?

Tip: Remember that the binomial coefficient $\binom{n}{k}$ is calculated as $\frac{n!}{k!(n-k)!}$, which represents the number of ways to choose $k$ elements from a set of $n$ elements.