Given that $n$ is an integer, for how many values of $n$ is $\frac{2 n^{2}-10 n-4}{n^{2}-4 n+3}$ an integer?
(A) 9
(B) 7
(C) 6
(D) 4
(E) 5

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-146.jpg?height=188&width=789&top_left_y=215&top_left_x=294)
Given that $n$ is an integer, for how many values of $n$ is $\frac{2 n^{2}-10 n-4}{n^{2}-4 n+3}$ an integer?
(A) 9
(B) 7
(C) 6
(D) 4
(E) 5

## Solution

We start by dividing $n^{2}-4 n+3$ into $2 n^{2}-10 n-4$.

$$
\begin{array}{r}
2 \\
n ^ { 2 } - 4 n + 3 \longdiv { 2 n ^ { 2 } - 1 0 n - 4 } \\
\frac{2 n^{2}-8 n+6}{-2 n-10}
\end{array}
$$

This allows us to write the original expression in the following way,

$$
\frac{2 n^{2}-10 n-4}{n^{2}-4 n+3}=2+\frac{-2 n-10}{n^{2}-4 n+3}=2-\frac{2 n+10}{n^{2}-4 n+3}
$$

The original question comes down to the consideration of $\frac{2 n+10}{n^{2}-4 n+3}$ and when this expression is an integer. This rational expression can only assume integer values when, $2 n+10 \geq n^{2}-4 n+3$ (the numerator must be greater than the denominator) and when $2 n+10=0$.

Or, $n^{2}-6 n-7 \leq 0$ and $n=-5$

or, $\quad(n-7)(n+1) \leq 0$

$-1 \leq n \leq 7$.

This means that we only have to consider values of $n,-1 \leq n \leq 7, n \in Z$ and $n=-5$. Also note that since $n^{2}-4 n+3=(n-1)(n-3)$ we can remove $n=1$ and $n=3$ from consideration. We construct a table and check each value.

| $n$ | -5 | -1 | 0 | 2 | 4 | 5 | 6 | 7 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\frac{2 n+10}{(n-3)(n-1)}$ | 0 | +1 | $\frac{10}{3}$ | -14 | 6 | $\frac{5}{2}$ | $\frac{22}{15}$ | 1 |

From this table we can see that there are just four acceptable values of $n$ that produce an integer.

Note also that $\frac{2 n+10}{n^{2}-4 n+3}$ would also be an integer if $2 n+10=0$ and $n^{2}-4 n+3 \neq 0$. Thus $n=-5$ is a fifth value since the denominator $\neq 0$.

ANSWER: (E)