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(D) Given the polynomial equation $x^n+px+q=0$ with roots $\alpha_1, \alpha_2, \ldots, \alpha_n$,we can write it as: $x^n+px+q = (x-\alpha_1)(x-\alpha

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$n \alpha_n^{n-1}+p$

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(D) Given the polynomial equation $x^n+px+q=0$ with roots $\alpha_1, \alpha_2, \ldots, \alpha_n$,we can write it as: $x^n+px+q = (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n)$. Dividing both sides by $(x-\alpha_n)$,we get: $\frac{x^n+px+q}{x-\alpha_n} = (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_{n-1})$. Taking the limit as $x 	o \alpha_n$ on both sides: $\lim_{x 	o \alpha_n} \frac{x^n+px+q}{x-\alpha_n} = (\alpha_n-\alpha_1)(\alpha_n-\alpha_2)\ldots(\alpha_n-\alpha_{n-1})$. Using $L$'$H$ôpital's Rule on the left side: $\lim_{x 	o \alpha_n} \frac{\frac{d}{dx}(x^n+px+q)}{\frac{d}{dx}(x-\alpha_n)} = \lim_{x 	o \alpha_n} (nx^{n-1}+p) = n\alpha_n^{n-1}+p$. Thus,$(\alpha_n-\alpha_1)(\alpha_n-\alpha_2)\ldots(\alpha_n-\alpha_{n-1}) = n\alpha_n^{n-1}+p$.

If $\alpha_1, \alpha_2, \ldots, \alpha_n$ are the roots of $x^n+px+q=0$,then $(\alpha_n-\alpha_1)(\alpha_n-\alpha_2) \ldots (\alpha_n-\alpha_{n-1})=$

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