Find the appropriate value of $a$ s.t. $f(m,n)=g(m) h(n)$ For $f(m,n) = amn + 10m + 20n + 5$ is there a specific value for $a$ which enables us to write $f$ as product of a function of $m$ multiplied with a function of $n$, that is to write $f(m, n) = g(m) h(n)$?
 A: We can write that
$$f(m, n) = g(m) h(n)\,.$$
Since you never have a term of the form $m^k$ or $n^k$ for $k>1$, you can decide that
$$g(m) = \alpha m + \beta\,,$$
$$h(n) = \gamma n + \delta\,.$$
Multiplying these functions together, you get
$$f(m, n) = \underbrace{\alpha \gamma\vphantom{\gamma}}_{a} mn + \underbrace{\alpha \delta\vphantom{\gamma}}_{10} m + \underbrace{\beta \gamma\vphantom{\gamma}}_{20} n + \underbrace{\beta \delta\vphantom{\gamma}}_{5}\,.$$
Now, identifying each term with what you know is given in your equation, you obtain the following system:
$$\left\{
\begin{array}{rcl}
\alpha \gamma & = & a\,,\\
\alpha \delta & = & 10\,,\\
\beta \gamma  & = & 20\,,\\
\beta \delta  & = & 5\,.
\end{array}
\right.$$
All you need to do is then to solve this, yielding
$$\left\{
\begin{array}{rcl}
\alpha & \ne & 0\,,\\
\beta & = & \alpha/2\,,\\
\gamma  & = & 40/\alpha\,,\\
\delta  & = & 10/\alpha\,.
\end{array}
\right.$$
Now, the value of $\alpha$ isn't specified, but we still have that $a = \alpha \gamma$!
Using this last fact, we easily find that
$$a = \alpha \gamma = \alpha \frac{40}{\alpha} = 40\,.$$
That means that your original function is
$$f(m, n) = 40mn + 10m + 20n + 5\,.$$
