TITLE: What is known about the polynomial factorization of power series?
QUESTION [12 upvotes]: Some power series factorize;  $1+\sum_{n=1}^\infty x^n=\prod_{n=1}^\infty (1+x^{2^n})$ and $1+\sum_{n=1}^\infty x^{2n}/(2n+1)!=\prod_{x=1}^\infty (1+x^2/n^2\pi^2)$ for example; while others do not----in particular, $1+\sum_{n=1}^\infty x^n/n!$. Specifically the question is: what is known about necessary and sufficient conditions on the coefficients of the power series $1+\sum_{n=1}^\infty a_nx^n$ (assuming a positive radius of convergence) for it to factorize as $\prod_{n=1}^\infty (1+ b_nx+c_nx^2)$, where the coefficients $a_n$, $b_n$, and $c_n$ are real constants?
Remarks  The Weierstrass factorization theorem doesn't directly answer the question, since here we allow Taylor series of non-entire functions (e.g. the first example above) and exclude non-polynomial factors. The roots of the question are algebraic: loosely stated, under what conditions can the fundamental theorem of algebra be pushed to infinity? A positive radius of convergence is supposed because formal power series with no functional meaning make me uncomfortable, and admitting them might complicate the answer. I previously posted an unanswered version of this question on Math.StackExchange.
Edit  Thanks to juan and Douglas Zare for showing my error in the preamble of the question by citing the interesting factorization of the exponential series by Gingold et al. As Aaron Meyerowitz indicates, restriction to linear and quadratic factors leads respectively to quite different situations. Alexandre Eremenko answered the linear case. Allowing polynomial factors of unrestricted degree opens a wide vista. The rather simple power series $1$ includes among its polynomial product representations, valid for $|x| \lt r_0$, such forms as
$$\left(1-\dfrac{p(x)}{r} \right) \prod_{n=0}^\infty \left[1+\left(\dfrac{p(x)}{r} \right)^{2^n} \right],$$
where $r_0$ is an arbitrary positive constant, $p(x)$ is an arbitrary polynomial, and $r$ is any real constant strictly exceeding $|p(x)|$ whenever $|x|\leqslant r_0$. Myriad expansions like these may be woven into any polynomial product expansion of any power series. I find this vista daunting and so will stick to the specified quadratic case, which remains unanswered so far.

REPLY [3 votes]: Here is an observation which shows the delicacy of factorization: Your second product $$\prod_{n=1}^\infty (1+x^2/n^2\pi^2)=\sum_{n=0}^\infty x^{2n}/(2n+1)!=\frac{\sinh{x}}{x}$$ is absolutely convergent but is only conditionally convergent if it is written as $$\prod_{n=1}^\infty (1+ix/n\pi)(1-ix/n\pi).$$ So an order dependent factorization does not sound like the fundamental theorem of algebra.
We can leave complex numbers out entirely by shifting to a famous product of Euler (which is not an Euler product.) 
$$\prod_{n=1}^\infty (1-x^2/n^2\pi^2)=\sin{x}/x=\sum_{n=0}^\infty (-x^2)^{n}/(2n+1)!$$ Of course this was not rigorously justified at the time and I think Euler was perhaps content to leave it as $$\prod_{n=1}^\infty (1+x/n\pi)(1-x/n\pi).$$ The heuristic justification (one treatment is page 3 of this article) was that the product gives a function $f(x)$ with $f(0)=1$ and a single root of $f(n\pi)=0$ for non-zero integers $n$. Hence it should be $\sin{x}/x$ for which we do know the power series. 
In volume I of Polya's Mathematics and Plausible Reasoning it is related that the motivation was to find the exact (and unknown) value of $\sum 1/n^2$ which does converge to about $1.644934.$ Of course comparing the coefficient of $x^2$ gives $\sum 1/n^2=\pi^2/6.$ The conclusion is "(extremely) daring", says Polya,   but it does agree with the known value and the method yields other similarly valid looking results, so it seems acceptable.