The coefficient of x^5 in the expansion of (x | vedclass

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(C) Given expression: $(x^2 - x - 2)^5 = ((x - 2)(x + 1))^5 = (x - 2)^5(x + 1)^5$.Using the Binomial Theorem:$(x - 2)^5 = \sum_{k=0}^{5} \binom{5}{k}

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$-81$

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(C) Given expression: $(x^2 - x - 2)^5 = ((x - 2)(x + 1))^5 = (x - 2)^5(x + 1)^5$.Using the Binomial Theorem:$(x - 2)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} (-2)^k = \binom{5}{0}x^5 - 5\binom{5}{1}x^4(2) + 10\binom{5}{2}x^3(4) - 10\binom{5}{3}x^2(8) + 5\binom{5}{4}x(16) - \binom{5}{5}(32)$$= x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32$.$(x + 1)^5 = \binom{5}{0} + \binom{5}{1}x + \binom{5}{2}x^2 + \binom{5}{3}x^3 + \binom{5}{4}x^4 + \binom{5}{5}x^5$$= 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$.To find the coefficient of $x^5$,we multiply terms from the two expansions:$(1)(x^5) + (-10)(5x^4) + (40)(10x^3) + (-80)(10x^2) + (80)(5x) + (-32)(1)$$= 1 - 50 + 400 - 800 + 400 - 32 = -81$.

The coefficient of $x^5$ in the expansion of $(x^2 - x - 2)^5$ is

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