Let p(x) be a polynomial such that p(x) - p'( | vedclass

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(A) Given $p(x) - p'(x) = x^n$. Since $p(x)$ is a polynomial,let $p(x) = a_m x^m + a_{m-1} x^{m-1} + \dots + a_1 x + a_0$. Then $p'(x) = m a_m x^{m-1}

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$n!$

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(A) Given $p(x) - p'(x) = x^n$. Since $p(x)$ is a polynomial,let $p(x) = a_m x^m + a_{m-1} x^{m-1} + \dots + a_1 x + a_0$. Then $p'(x) = m a_m x^{m-1} + (m-1) a_{m-1} x^{m-2} + \dots + a_1$. Substituting into the equation: $p(x) = p'(x) + x^n$. Comparing the highest degree terms,we must have $m = n$ and $a_n = 1$. We can write $p(x) = (D - 1)^{-1} (-x^n)$,where $D = \frac{d}{dx}$. Using the operator expansion $(D - 1)^{-1} = -(1 - D)^{-1} = -(1 + D + D^2 + \dots + D^n)$. So,$p(x) = -(1 + D + D^2 + \dots + D^n) x^n$. $p(x) = -(x^n + n x^{n-1} + n(n-1) x^{n-2} + \dots + n!)$. Evaluating at $x = 0$,we get $p(0) = -n!$. However,checking the standard form $p(x) - p'(x) = x^n$,if we assume $p(x) = \sum_{k=0}^n a_k x^k$,then $a_0 - a_1 = 0 \implies a_0 = a_1$. By induction,$a_k = k! a_0$. For $x^n$,$a_n = 1$,so $1 = n! a_0 \implies a_0 = \frac{1}{n!}$. Wait,the correct relation is $p(x) = x^n + n x^{n-1} + n(n-1) x^{n-2} + \dots + n!$. Thus $p(0) = n!$.

Let $p(x)$ be a polynomial such that $p(x) - p'(x) = x^n$,where $n$ is a positive integer. Then,$p(0)$ equals

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