Review question

# What is the remainder when $p_n (x)$ is divided by $p_{n-1} (x)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5276

## Solution

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

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

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

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

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).