# 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 $\alpha \in \mathbb{C}$

#### Solutions Collecting From Web of "When does the product of two polynomials = $x^{k}$?"

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)

It’s true over any domain $\rm D$ since then $x$ is prime (via $\rm D[x]/x = D$ a domain), and products of primes always factor uniquely in domains. It fails over non-domains, e.g. $\rm x = (2x+3)(3x+2) \in \mathbb Z/6[x].$

The answer just occured to me. The roots of $f$ and $g$ must be be at 0.

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 (f/x) 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.