What I mean to ask is this:


given an irreducible **cubic** polynomial $P(X)\in \mathbb{Z}[X]$ is there always a **quadratic** $Q(X)\in \mathbb{Z}[X]$ such that $P(Q)$ is reducible (as a polynomial, and then necessarily the product of 2 irreducible cubic polynomials)?


I did quite some testing and always found a $Q$ that does the job. For example:


$P=aX^3+b,\quad Q=-abX^2,\quad P(Q)=-b(a^2bX^3-1)(a^2bX^3+1)$


$P=aX^3-x+1,\quad Q=-aX^2+X,\quad P(Q)=-(a^2X^3-2aX^2+X-1)(a^2X^3-aX^2+1)$


and a particular hard one to find:


$P=2X^3+X^2-X+4,\quad Q=-8X^2+5X+1,\quad P(Q)=(16X^3-18X^2+X+3)(64X^3-48X^2-11x-2)$


Could there be a formula for $Q$ that works for all cases?


It feels to me that this may have a really basic Galois theoretic proof or explanation, but I can't figure it out.


**Update.** Maybe a general formula for $Q$ is close. For $P=aX^3+cX+d$ taking $Q=-adX^2+cX$ works.



Does any cubic polynomial become reducible through composition with some quadratic?
You should refer to Lemma 10 (page-233) in [this](https://eudml.org/doc/204826)  paper by Schinzel where he proves that for any polynomial $F(x)$ of degree $d$ we have a polynomial $G(x)$ of degree $d-1$ such that their composition is reducible.


