We are asked to expand the binomial $(3x^2 + x^3)^5$ and identify the fourth term when ordered in descending powers of $x$.

To solve this, we will use the Binomial Theorem, which states:

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

Here:

• $a = 3x^2$
• $b = x^3$
• $n = 5$

The general term of the expansion will be given by:

$T_k = \binom{5}{k} (3x^2)^{5-k} (x^3)^k$

We will calculate the first few terms to find the fourth one.

1. For $k = 0$: $T_0 = \binom{5}{0} (3x^2)^5 (x^3)^0 = 1 \cdot 243x^{10} = 243x^{10}$

2. For $k = 1$: $T_1 = \binom{5}{1} (3x^2)^4 (x^3)^1 = 5 \cdot 81x^8 \cdot x^3 = 5 \cdot 81x^{11} = 405x^{11}$

3. For $k = 2$: $T_2 = \binom{5}{2} (3x^2)^3 (x^3)^2 = 10 \cdot 27x^6 \cdot x^6 = 10 \cdot 27x^{12} = 270x^{12}$

4. For $k = 3$: $T_3 = \binom{5}{3} (3x^2)^2 (x^3)^3 = 10 \cdot 9x^4 \cdot x^9 = 10 \cdot 9x^{13} = 90x^{13}$

Since the polynomial is ordered in decreasing powers of $x$, the terms are:

$90x^{13}, 270x^{12}, 405x^{11}, 243x^{10}, \ldots$

The fourth term is $405x^{11}$.

Thus, the correct answer is option b) 405x^{11}.

Would you like further clarification or details?

Here are 5 related questions:

1. How does the Binomial Theorem work for negative exponents?
2. What are binomial coefficients and how are they calculated?
3. Can the expansion of binomials be used for more complex expressions like trinomials?
4. What happens if we substitute $x = 1$ into the binomial expansion?
5. How does Pascal’s Triangle relate to binomial expansions?

Tip: The Binomial Theorem simplifies polynomial expansions and helps in finding specific terms without full expansion.