Express polynomial in terms of one coefficient Take the polynomial
$$2 a b n^2 - a^2 b^2 n$$
where $n$ is the variable, and coefficients $a$ and $b$ are integers.
Is it possible to derive an expression for $b$ in terms of $a$ and $n$, or an expression for $a$ in terms of $b$ and $n$?
UPDATE:
It seems I wasn't as clear as I thought. My bad. Here is what I meant to say... Given the expression $2 a b n^2 - a^2 b^2 n$, is it possible to completely factor out either $a$ or $b$ so that, for two functions $f$ and $g$, the expression $2 a b n^2 - a^2 b^2 n$ can be rewritten as
$$2 a b n^2 - a^2 b^2 n=f(b)g(a,n)$$
or
$$2 a b n^2 - a^2 b^2 n=f(a)g(b,n)$$
Hopefully that's clearer. You were right to pull me up. Thank you.
 A: Suppose there exist function $f$ and $g$ such that
$$2 a b n^2 - a^2 b^2 n=f(a)g(b,n).$$
First note that if $n$ is nonzero, then for any nonzero $b$ with $b\neq2n$ we have
$$f(1)g(b,n)=2bn^2-b^2n=bn(2n-b)\neq0,$$
so in particular $g(b,n)\neq0$. Then also $bn$ is nonzero and $b\neq 2bn$, so $g(b,bn)\neq0$ and
$$f(2n)g(b,bn)
=2\cdot(2n)\cdot b\cdot(bn)^2-(2n)^2\cdot b^2\cdot(bn)=0,$$
which shows that $f(2n)=0$ for all nonzero $n$. But for all nonzero $n$ we also have
$$f(n)g(n,n^2)=2\cdot n\cdot n\cdot n^4-n^2\cdot n^2\cdot n^2=n^6\neq0,$$
a contradiction. Hence no such functions $f$ and $g$ exist.
A: One thing to do with a question like this is to just try some numbers. 
Here are values of the polynomial for a few values of $a,$ $b,$ and $n.$
\begin{array}{cccc}
a&b&n&2abn^2-a^2b^2n\\
1&1&1&1\\
1&1&2&6\\
1&2&1&0\\
1&2&2&8
\end{array}
From the third row, we see that if the polynomial can be expressed as the product $f(b)g(a,n),$ either $f(2)=0$ or $g(1,1)=0.$
But the first row requires that $g(1,1)\neq0,$
and the fourth row requires that $f(2)\neq0.$
There is no way to reconcile these three rows. 
The polynomial is symmetric in $a$ and $b,$
so we know even without further checking that $f(a)g(b,n)$ also will not work. 
A: Above equations shown below:
$2 a b n^2 - a^2 b^2 n=w=f(b)g(a,n)$  -----$(1)$
$2 a b n^2 - a^2 b^2 n=w=f(a)g(b,n)$  ------$(2)$
Equations (1) & (2) has solutions:
$(a,b,n,w)=[(p^2),(q^2),(p^2q^2),(p^6q^6)]$
$g(a,n)=(an^2)=(p^6q^4)$
$g(b,n)=(bn^2)=(p^4q^6)$
$f(a)=(a)=(p^2)$
$f(b)=(b)=(q^2)$
A: Given only $2abn^2-a^2b^2n$, let's assume standard form $2a^2b^2n-2abn^2+0=0$. Then we can use the quadratic equation to solve for either $a$ or $b$, here solved for $a$.
$$a=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2bn^2\pm\sqrt{(2bn^2)^2-4(b^2n)(0)}}{2(2b^2n)}=\frac{2bn^2\pm 2bn^2}{2(2b^2n)}=\frac{4bn(n)\lor 0}{4bn(b)}=\frac{n}{b}\lor0$$
Since the powers of $a$ and $b$ are the same, $b=\frac{n}{a}\text{ or }0$ and this answers the original question. I'm not sure what you want to do with the part about f(a)g(...
