Let's solve both parts of this problem using the binomial theorem.

Part A: Coefficient of the fifth term of $(2x + 3y)^6$

The general form of the binomial expansion for $(a + b)^n$ is:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

For this problem:

• $a = 2x$
• $b = 3y$
• $n = 6$
• We want the fifth term, so $k = 4$ (since the term number is $k+1$).

Thus, the fifth term is:

$T_5 = \binom{6}{4} (2x)^{6-4} (3y)^4 = \binom{6}{4} (2x)^2 (3y)^4$

Now, calculate each part:

• $\binom{6}{4} = 15$
• $(2x)^2 = 4x^2$
• $(3y)^4 = 81y^4$

So, the fifth term is:

$T_5 = 15 \cdot 4x^2 \cdot 81y^4 = 4860x^2 y^4$

The coefficient is 4860.

Part B: Coefficient of the fourth term of $(4x - 3y)^5$

Here, we apply the binomial expansion again. For this part:

• $a = 4x$
• $b = -3y$
• $n = 5$
• We want the fourth term, so $k = 3$ (since the term number is $k+1$).

Thus, the fourth term is:

$T_4 = \binom{5}{3} (4x)^{5-3} (-3y)^3 = \binom{5}{3} (4x)^2 (-3y)^3$

Now, calculate each part:

• $\binom{5}{3} = 10$
• $(4x)^2 = 16x^2$
• $(-3y)^3 = -27y^3$

So, the fourth term is:

$T_4 = 10 \cdot 16x^2 \cdot (-27y^3) = -4320x^2 y^3$

The coefficient is -4320.

Summary:

• Part A: The coefficient of the fifth term is 4860.
• Part B: The coefficient of the fourth term is -4320.

Would you like more details or have any questions? Here are five related questions to deepen your understanding:

1. How does the binomial theorem work for higher powers of binomials?
2. Can you find the general term for $(x + y)^n$ using the binomial formula?
3. What is the role of binomial coefficients in the expansion of $(a + b)^n$?
4. How does the sign change when you expand binomials like $(a - b)^n$?
5. How would the coefficients change if the exponents in the binomial expansion increase?

Tip: The binomial coefficient $\binom{n}{k}$ can be calculated as $\frac{n!}{k!(n-k)!}$, where $n!$ denotes the factorial of $n$.