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.
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.
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(...