- Note that $n$ is a positive integer. For $n=1$,
$\begin{array}{rcl}
m + 2(1) & = & 5 \\
m & = & 3
\end{array}$For $n=2$,
$\begin{array}{rcl}
m + 2(2) & = & 5 \\
m & = & 1
\end{array}$For $n=3$,
$\begin{array}{rcl}
m + 2(3) & = & 5 \\
m & = & -1
\end{array}$Hence, the larger the value of $n$, the smaller the value of $m$. Therefore, all positive integers $m$ are $1$ and $3$.
-
$\begin{array}{rcl}
2x^2 + 5x + k & \equiv & (2x+m)(x+n) \\
2x^2 + 5x + k & \equiv & 2x^2 +mx +2nx + mn \\
2x^2 + 5x + k & \equiv & 2x^2 + ( m+2n) x + mn
\end{array}$By comparing the coefficients of both sides, we have
$\left\{ \begin{array}{ll}
m+2n = 5 & \ldots \unicode{x2460} \\
mn = k & \ldots \unicode{x2461}
\end{array} \right.$Consider equation $\unicode{x2460}$, by the result of (a), we have $\left\{ \begin{array}{l} n=1 \\ m=3 \end{array}\right.$ or $\left\{ \begin{array}{l} n=2 \\ m=1 \end{array} \right.$. Substitute the first set of values into $\unicode{x2461}$, we have
$\begin{array}{rcl}
k & = & (3)(1) \\
& = & 3
\end{array}$Substitute the second set of values into $\unicode{x2461}$, we have
$\begin{array}{rcl}
k & = & (2)(1) \\
& = & 2
\end{array}$Therefore, the values of $k$ are $2$ and $3$.
2008-I-03
Ans: (a) $1$ and $3$ (b) $2$ and $3$
