# How to find the remainder of polynomial division?

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?

• I am having trouble understanding your definition of $p_n(x)$. Are you multiplying all of $(x-2)(x-3)\cdots (x-n)$? – Joel May 16 '15 at 22:46
• No, I forgot to add a addition symbol before (x-n): $$p_n(x) = (x-1)+(x-2)+...+(x-n)$$ – Felipe ZC May 16 '15 at 22:47
• I see. That makes things simpler. – Joel May 16 '15 at 22:49

## 3 Answers

$p_n\,$ 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\!:\$ $\color{#c00}d = p_n(\frac{n}2) = p_{n-1}(\frac{n}2)+(\frac{n}2\!-\!n) = -\frac{n}2$

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.

• I dont understand what you mean by "matching the lead coefficient", why do you multiply by $n/n-1$? – Felipe ZC May 16 '15 at 23:34
• You should write out $p_{n-1}$ and try what I suggested. When you subtract the result from $p_n(x)$ it will cancel with $nx$. – Joel May 16 '15 at 23:40
• I am just describing the long division algorithm in words here. – Joel May 16 '15 at 23:40

Hint: Simplify $p_n(x)$: $$p_n(x) = (x-1)+(x-2)+\cdots+(x-n) = nx-{n(n+1)\over2}$$ and $$p_{n-1} = (n-1)x-{(n-1)n\over2}$$

• How would you divide these two expressions? – Felipe ZC May 16 '15 at 23:34