When does the product of two polynomials = $x^{k}$? Suppose $f$ and $g$ are are two polynomials with complex coefficents (i.e $f,g \in \mathbb{C}[x]$).
Let $m$ be the order of $f$ and let $n$ be the order of $g$.
Are there some general conditions where  
$fg= \alpha x^{n+m}$
for some non-zero $\alpha \in \mathbb{C}$ 
 A: Polynomials over $\mathbb{C}$ (in fact, over any field) are a Unique Factorization Domain (see http://en.wikipedia.org/wiki/Unique_factorization_domain); since $x$ is an irreducible, the only way for that to happen is for $f=ax^m$ and $g=bx^n$, with $ab=\alpha$. 
(If you don't want to bring in the sledgehammer of unique factorization, you can just do it explicitly: look at the lowest nonzero term in $f$ and the lowest nonzero term in $g$; their product will be the lowest nonzero term in $fg$, hence must be of degree $m+n$. Since the degree of the lowest nonzero term of $f$ is at most $m$ and the one of $g$ is at most $n$, you have that they must be exactly of degree $m$ and $n$, respectively, and you get the result)
A: We don't need the strong property of UFD. If $\rm D$ is a domain $\rm D$ then $\rm x$ is prime in $\rm D[x]$ (by $\rm D[x]/x \cong D$ a domain), and products of primes factor uniquely in every domain (same simple proof as in $\Bbb Z$). In particular, the only factorizations of the prime power $\rm x^i$ are $\rm \,x^j x^k,\ i = j+k\ $ (up to associates as usual). This fails  over non-domains, e.g. $\,\rm x = (2x+3)(3x+2) \in \mathbb Z/6[x].$
A: The answer just occured to me.   The roots of $f$ and $g$ must be be at 0. 
A: Yes, the intuitively evident ones: all other terms in $f$ and $g$ must vanish.  To see this, note that the product of the constant terms of $f$ and $g$ equals the constant term of $fg$, which is zero, whence at least one of these polynomials is multiple of $x$.  Without any loss of generality assume it is $f$.  Then
$$fg = x\, \left(\frac{f}{x}\right) g,$$
implying $(f/x) g$ is a multiple of $x^{n+m-1}$.  By induction this reduces us to the case $n+m=0$, which is trivial (because $f$ and $g$ then have no other terms). QED.
