<p>The coefficient of \(x^3\) in the expansion of \((2 - x + 3x^2)^5\) is ______.</p>
Step-by-Step Solution
Key Concept: Rewrite the trinomial as a binomial by grouping terms, then use the multinomial theorem or systematic expansion by considering all ways to select terms that produce x³. Alternatively, expand (2 - x + 3x²)⁵ by treating it as (2 + (-x + 3x²))⁵ and apply binomial theorem with careful coefficient tracking.
<p><strong>Step 1:</strong> Identify all ways to form x³ from five factors of (2 - x + 3x²):</p><p>We need terms where the powers of x sum to 3. If we choose: 2 appears a times, -x appears b times, and 3x² appears c times, then a + b + c = 5 and b + 2c = 3.</p><p><strong>Step 2:</strong> Find valid (a, b, c) combinations:</p><ul><li><strong>Case 1:</strong> c = 0, b = 3, a = 2 → Choose -x three times and 2 twice: $\binom{5}{3,2,0}·2^2·(-1)^3·1 = \frac{5!}{3!2!}·4·(-1) = 10·(-4) = -40$</li><li><strong>Case 2:</strong> c = 1, b = 1, a = 3 → Choose 3x² once, -x once, and 2 thrice: $\binom{5}{3,1,1}·2^3·(-1)^1·3 = \frac{5!}{3!1!1!}·8·(-1)·3 = 20·(-24) = -480$</li></ul><p><strong>Step 3:</strong> Sum all cases: $-40 + (-480) = -520$</p><p>∴ Answer: <strong>-520</strong></p>
Correct Answer: -520