Problem 1907

Solution:
Let $f(x)=d_{9} x^{9}+d_{8} x^{8}+\cdots+d_{1} x+d_{0}$. If $d_{0}=0$, then $f(x)$ is divisible by $x$ and thus reducible, so we may ignore all such polynomials. The remaining polynomials all have nonzero leading and constant coefficients.
For any polynomial $p(x)$ of degree $n$ with nonzero leading and constant coefficients, say $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, define $\bar{p}(x)$ to be the reversed polynomial $a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$. Observe that $\bar{p}(x)$ also has degree $n$ and furthermore, $\bar{p}(x)=x^{n}\left(a_{0}+a_{1}\left(\frac{1}{x}\right)+\cdots+a_{n-1}\left(\frac{1}{x}\right)^{n-1}+a_{n}\left(\frac{1}{x}\right)^{n}\right)=x^{n} p\left(\frac{1}{x}\right)$.
Consider pairing each $f(x)$ with $\bar{f}(x)$ whenever $f(x) \neq \bar{f}(x)$. If $f(x)$ is reducible, it can be factored as $f(x)=g(x) h(x)$ where $\operatorname{deg} g, \operatorname{deg} h \geq 1$. Because the leading and constant coefficients of $f(x)$ are nonzero, so are the leading and constant coefficients of $g(x)$ and $h(x)$. Hence $\bar{g}(x)$ and $\bar{h}(x)$ are well defined with $\operatorname{deg} \bar{g}=\operatorname{deg} g \geq 1$ and $\operatorname{deg} \bar{h}=\operatorname{deg} h \geq 1$. Furthermore,
$$\bar{f}(x)=x^{9} f\left(\frac{1}{x}\right)=x^{9} g\left(\frac{1}{x}\right) h\left(\frac{1}{x}\right)=\left(x^{\operatorname{deg} g} g\left(\frac{1}{x}\right)\right)\left(x^{\operatorname{deg} h} h\left(\frac{1}{x}\right)\right)=\bar{g}(x) \bar{h}(x)$$
Thus $\bar{f}(x)=\bar{g}(x) \bar{h}(x)$ is a factorization of $\bar{f}(x)$ into two non-constant polynomials, so $\bar{f}(x)$ is also reducible. Therefore $f(x)$ is irreducible if and only if $\bar{f}(x)$ is irreducible, so considering each pair, there are an even number of irreducible polynomials with $f(x) \neq \bar{f}(x)$.
Finally, note that if $f(x)=\bar{f}(x)$, then $d_{i}=d_{9-i}$ for each $i$. In such a case, we have $f(-1)=\left(d_{0}-d_{9}\right)+\left(d_{2}-d_{7}\right)+\left(d_{4}-d_{5}\right)+\left(d_{6}-d_{3}\right)+\left(d_{8}-d_{1}\right)=0$, so by the Factor Theorem, $(x+1)$ is a factor of $f(x)$. Therefore these remaining polynomials are all reducible.