In the expansion of (x - 2)^8, what is the coefficient of x^5?

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Q: In the expansion of (x - 2)^8, what is the coefficient of x^5?

Solution: The coefficient of x^5 is C(8,5) * (-2)^3 = 56 * (-8) = -448.

Steps: 10

Step 1: Identify the expression to expand, which is (x - 2)^8.
Step 2: 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 3: In our case, a = x, b = -2, and n = 8.
Step 4: We want the coefficient of x^5, which means we need to find the term where the power of x is 5.
Step 5: To find this term, we set n - k = 5, which means k = 8 - 5 = 3.
Step 6: Calculate the binomial coefficient C(8, 3), which is the number of ways to choose 3 from 8.
Step 7: C(8, 3) = 8! / (3! * (8-3)!) = 56.
Step 8: Now, calculate (-2)^3, which is -2 * -2 * -2 = -8.
Step 9: Multiply the coefficient C(8, 3) by (-2)^3: 56 * (-8) = -448.
Step 10: The coefficient of x^5 in the expansion of (x - 2)^8 is -448.