To expand \((2x + 3y)^5\) using the binomial theorem, we use the general formula for the binomial expansion:
\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]
Here, \(a = 2x\), \(b = 3y\), and \(n = 5\).
So, the expansion is:
\[
(2x + 3y)^5 = \sum_{k=0}^{5} \binom{5}{k} (2x)^{5-k} (3y)^k
\]
Now we will write down each term in the expansion:
1. For \(k = 0\):
\[
\binom{5}{0} (2x)^5 (3y)^0 = 1 \cdot 32x^5 \cdot 1 = 32x^5
\]
2. For \(k = 1\):
\[
\binom{5}{1} (2x)^4 (3y)^1 = 5 \cdot 16x^4 \cdot 3y = 240x^4y
\]
3. For \(k = 2\):
\[
\binom{5}{2} (2x)^3 (3y)^2 = 10 \cdot 8x^3 \cdot 9y^2 = 720x^3y^2
\]
4. For \(k = 3\):
\[
\binom{5}{3} (2x)^2 (3y)^3 = 10 \cdot 4x^2 \cdot 27y^3 = 1080x^2y^3
\]
5. For \(k = 4\):
\[
\binom{5}{4} (2x)^1 (3y)^4 = 5 \cdot 2x \cdot 81y^4 = 810xy^4
\]
6. For \(k = 5\):
\[
\binom{5}{5} (2x)^0 (3y)^5 = 1 \cdot 1 \cdot 243y^5 = 243y^5
\]
Now, combine all these terms to get the expanded form:
\[
(2x + 3y)^5 = 32x^5 + 240x^4y + 720x^3y^2 + 1080x^2y^3 + 810xy^4 + 243y^5
\]
This is the binomial expansion of \((2x + 3y)^5\).
