Coefficient of x^2 in the expansion of (x^2+x | vedclass

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(C) The given expression is $(x^2+x-2)^5$. We can factorize the quadratic expression $x^2+x-2$ as $(x+2)(x-1)$. So,the expression becomes $((x+2)(x-1)

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(C) The given expression is $(x^2+x-2)^5$. We can factorize the quadratic expression $x^2+x-2$ as $(x+2)(x-1)$. So,the expression becomes $((x+2)(x-1))^5 = (x+2)^5(x-1)^5$. Using the binomial theorem,$(x+2)^5 = \sum_{k=0}^{5} \binom{5}{k} x^k 2^{5-k}$ and $(x-1)^5 = \sum_{j=0}^{5} \binom{5}{j} x^j (-1)^{5-j}$. We need the coefficient of $x^2$ in the product of these two expansions. The product is $(\binom{5}{0}2^5 + \binom{5}{1}2^4 x + \binom{5}{2}2^3 x^2 + \dots) 	imes (\binom{5}{0}(-1)^5 + \binom{5}{1}(-1)^4 x + \binom{5}{2}(-1)^3 x^2 + \dots)$. Let $A = (32 + 80x + 80x^2 + \dots)$ and $B = (-1 + 5x - 10x^2 + \dots)$. The coefficient of $x^2$ is obtained by: $(32 	imes -10) + (80 	imes 5) + (80 	imes -1) = -320 + 400 - 80 = 0$. Thus,the coefficient of $x^2$ is $0$.

Coefficient of $x^2$ in the expansion of $(x^2+x-2)^5$ is

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