Do even degree polynomials have even degree factors (no conjugates)?

Let $R:=\Bbb{Z}[X^2]\subset\Bbb{Z}[X]$ be the subring of 'even degree' polynomials. Let $P\in R$ factor in $\Bbb{Z}[X]$ as $$P=\prod_{i=1}^nQ_i,$$ where the $Q_i$ are irreducible. Because $P\in R$ we have $P(X)=P(-X)$ and so $$\prod_{i=1}^nQ_i(-X)=P(-X)=P(X)=\prod_{i=1}^nQ_i(X),$$ so by unique factorization, for each $i$ there exists some $j$ such that $Q_i(-X)=Q_j(X)$. In particular the coefficients of $Q_i$ and $Q_j$ are congruent mod $2$, and so for the polynomials $$F:=\frac{Q_i+Q_j}{2}\qquad\text{ and }\qquad G:=\frac{Q_i-Q_j}{2},$$ we have $F,G\in\Bbb{Z}[X]$ and $Q_i=F+G$ and $Q_j=F-G$. So $Q_i$ and $Q_j$ are a conjugate pair, unless $G=0$ which is the case if and only if $Q_i=Q_j$.

So if $P$ does not have any conjugate pairs we must have $Q_i(X)=Q_i(-X)$ for all $i$, meaning that $Q_i\in R$ for all $i$, so indeed all factors of $P$ are 'even degree' as well.

Suppose $f(x)$ has even degree and let $f(x)=p(x)q(x)$ where $p(x)$ is an irreducible factor of $f(x)$. Then $f(-x)=p(-x)q(-x)$ and so $p(-x)$ is a factor of $f(x)$ as well. If $p(-x) \neq p(x)$, then $p(-x)$ divides $q(x)$ and so $f(x)=p(x)p(-x)g(x)$. Clearly $p(x)$ and $p(-x)$ are conjugate pairs.