What is the remainder when $p_n (x)$ is divided by $p_{n-1} (x)$?

What is the remainder when $p_n (x)$ is divided by $p_{n-1} (x)$?

Question

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 $p_{n-1} (x)$?

1. $\dfrac{n}{2}$;

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1. $\dfrac{n+1}{2}$;

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1. $\dfrac{n^2+n}{2}$;

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1. $\dfrac{-n}{2}$.

Suggestion

The terms in $p_n (x)$ follow a sequence - can we sum this?

What value of $x$ makes $p_{n-1}(x)$ zero?

Solution

We can use the formula for the sum of an arithmetic progression to calculate $p_n(x) = nx - \frac{n(n+1)}{2}$ and $p_{n-1}(x) = (n-1)x - \frac{n(n-1)}{2}.$

Approach 1

We can carry out long division, by noting that the

$x$

term of

$p_{n-1}(x)$

divides

$\dfrac{n}{n-1}$

times into the

$x$

term of

$p_n(x)$

, so that the remainder is

$$\begin{align*} p_n(x) - \dfrac{n}{n-1}p_{n-1}(x) &{}= \left(nx - \dfrac{n(n+1)}{2}\right) - \dfrac{n}{n-1} \left((n-1)x - \dfrac{n(n-1)}{2}\right)\\ &{}= -\dfrac{n(n+1)}{2} + \dfrac{n^2}{2}\\ &{}= \dfrac{-n}{2}, \end{align*}$$

and the answer is (d).

Approach 2

We can use the Remainder Theorem. We have that $p_{n-1}(x)=(n-1)\bigl(x-\tfrac{1}{2}n\bigr),$ and so $p_{n-1}(x) =0$ when $x = \dfrac{n}{2}$ (we can ignore the $n-1$ factor as it will not change the remainder).

So the remainder is equal to $p_n\left(\dfrac{n}{2}\right) = -\dfrac{n}{2}$, as before.

The answer is (d).