What is the coefficient of x^3 in the expansion of (2x + 3)^5?

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Q: What is the coefficient of x^3 in the expansion of (2x + 3)^5?

Solution: Using the binomial theorem, the coefficient of x^3 is given by C(5,3) * (2)^3 * (3)^(5-3) = 10 * 8 * 9 = 720.

Steps: 10

Step 1: Identify the expression to expand, which is (2x + 3)^5.
Step 2: Recognize that we need to find the coefficient of x^3 in this expansion.
Step 3: Use the binomial theorem, which states that (a + b)^n = Σ [C(n, k) * a^(n-k) * b^k] for k = 0 to n.
Step 4: In our case, a = 2x, b = 3, and n = 5.
Step 5: We want the term where the power of x is 3, which means we need to set k = 5 - 3 = 2.
Step 6: Calculate C(5, 2), which is the number of ways to choose 2 from 5. C(5, 2) = 5! / (2!(5-2)!) = 10.
Step 7: Calculate (2)^3, which is the coefficient of x^3. (2)^3 = 8.
Step 8: Calculate (3)^(5-3), which is (3)^2 = 9.
Step 9: Multiply the results from steps 6, 7, and 8: 10 * 8 * 9.
Step 10: The final result is 720, which is the coefficient of x^3.