To determine the coefficient of $x^5$ in the expansion of $(2x + 3)^6$, we apply the binomial theorem, which states:

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

Here:

• $a = 2x$,
• $b = 3$,
• $n = 6$.

The general term in the expansion is:

$T_k = \binom{6}{k} (2x)^{6-k} (3)^k$

Step 1: Finding the exponent of $x$

The term involves $x$ raised to the power of $6-k$ because $x$ appears in $(2x)^{6-k}$.

For $x^5$, set the exponent of $x$ equal to $5$:

$6 - k = 5 \implies k = 1$

Step 2: Substitute $k = 1$ into the general term

Now, substituting $k = 1$ into the general term $T_k$:

$T_1 = \binom{6}{1} (2x)^{6-1} (3)^1$

Simplify each component:

• $\binom{6}{1} = 6$,
• $(2x)^5 = 2^5 \cdot x^5 = 32x^5$,
• $3^1 = 3$.

Thus:

$T_1 = 6 \cdot 32x^5 \cdot 3$

Step 3: Calculate the coefficient

Combine the constants:

$6 \cdot 32 \cdot 3 = 576$

Final Answer:

The coefficient of $x^5$ is 576.

Do you want a further breakdown of any step or additional examples?

Related Questions:

1. What is the general term in a binomial expansion?
2. How do you determine the power of $x$ in a specific term?
3. What are the properties of binomial coefficients $\binom{n}{k}$?
4. How does the binomial theorem apply when terms involve powers and constants?
5. What happens to the coefficients when the binomial contains fractions or negative terms?

Tip:

Always identify the term's power of $x$ and solve for $k$ before simplifying the coefficient.