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We say that $T(X) \in \mathbb{Q}[X]$ is a trinomial if there exist $A,B,C \in \mathbb{Q}$ such that $T(X) = AX^n + BX^m + C$ for some $n \geq m \in \mathbb{N}$.

Is it true that for each irreducible $g(X) \in \mathbb{Q}[X]$ there exists some nonzero $h(X) \in \mathbb{Q}[X]$ such that $g(X)h(X)$ is a trinomial?

If no, then what are some necessary/sufficient conditions on $g$ for the existence of such $h$?

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    $\begingroup$ All non-zero multiples of $(X-1)^k$ require at least $k+1$ non-zero coefficients. A necessary condition on $g$ is that it does not have a root $\alpha \neq 0$ of multiplicity $> 2$. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 12, 2015 at 12:17
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    $\begingroup$ How many real zeros can a trinomial have? $\endgroup$
    – Gerald Edgar
    Commented Aug 12, 2015 at 12:19
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    $\begingroup$ @VesselinDimitrov thanks! I thought about irreducible polynomials. I will edit accordingly. $\endgroup$
    – Pablo
    Commented Aug 12, 2015 at 12:27
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    $\begingroup$ What's the answer over finite fields? $\endgroup$
    – Dror Speiser
    Commented Aug 12, 2015 at 13:00
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    $\begingroup$ There are a few other splitting constraints as well; e.g., a trinomial has only a bounded number of $\mathbb{Q}_p$-roots, for every fixed $p$. (Also Khovanskii's theory of fewnomials could be something worth looking up for generalizations of this kind of question.) But it would be quite interesting to show that "almost all" degree-$d$ polynomials in $\mathbb{Z}[X]$ (or in $\mathbb{R}[X]$) require at least $d+1$ non-zero coefficients for all their multiples. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 12, 2015 at 13:10

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To summarize some of the discussion in the comments: a trinomial can have at most four distinct real roots. A random polynomial of degree $d$ (random = all coefficients are iid, there are other models) has $\Omega(\log d)$ real roots. Which means that the probability that a random polynomial divides a trinomial goes to zero (exponentially fast with degree, though this is a little harder to prove).

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    $\begingroup$ Should the notation be $\Omega(\log{d})$, or $\asymp \log{d}$? With $O(\cdot)$ one could infer that only an upper bound is stated, while actually here it is the lower bound that matters. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 12, 2015 at 16:56
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    $\begingroup$ @VesselinDimitrov Correct, it is actually asymptotic to constant times log. $\endgroup$
    – Igor Rivin
    Commented Aug 12, 2015 at 19:00
  • $\begingroup$ @VesselinDimitrov OK, fixed... $\endgroup$
    – Igor Rivin
    Commented Aug 12, 2015 at 19:10

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