Im trying to solve this problem but I do not understand what the question is asking:
Let $n\ge 2$ be an integer and $ p_n(x) $ be the polynomial: $$ p_n(x) = (x-1)+(x-2)+\cdots+(x-n) $$
What is the remainder when $ p_n(x)$ is divided by $ p_{n-1}(x)$?
I don't understand the problem, how would I go about solving it?
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Hint: Simplify $p_n(x)$: $$p_n(x) = (x-1)+(x-2)+\cdots+(x-n) = nx-{n(n+1)\over2}$$ and $$p_{n-1} = (n-1)x-{(n-1)n\over2}$$
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We can write $p_n(x)$ as $$p_{n}(x) = n x - (1+2+3+\cdots + n) = nx - \frac{n(n+1)}{2}$$
When we do long division, part of the algorithm is to match the lead coeficient, so here we want to change the leading coefficent of $p_{n-1}$ to match with $p_n(x)$.
Thus you need to compute $\frac{n}{n-1} \cdot p_{n-1}(x)$, which will have a leading term of $nx$ that matches $p_n(x)$.
After you match leading terms, you subtract $\frac{n}{n-1} \cdot p_{n-1}(x)$ from $p_n(x)$. The result is a constant. This is your remainder.
$p_n\,$ is linear so $\, p_n(x) = c\, p_{n-1}(x) + \color{#c00}d.\,$ $\,p_{n-1}(x)\,$ has root = midpoint of $\,1,2,\ldots,n\!-\!1 =$ $\, \frac{1+(n-1)}2 = \frac{n}2.\ $ Evaluating at $\,x=\frac{n}2\!:\ $ $ \color{#c00}d = p_n(\frac{n}2) = p_{n-1}(\frac{n}2)+(\frac{n}2\!-\!n) = -\frac{n}2$