$\newcommand\S{\mathcal S}$ Let $p$ be a prime, and suppose that $0.9(p-1)<q<p-1$. Suppose, furthermore, that $H,B,T\in\mathbb F_p[x]$ satisfy $\max\{\deg H,\deg T\}\le q-1$ and $\deg B\le p-1-q$. What can be said about $B$ given that the polynomial $$ \S(x) := x^p H(x) + x^q B(x) + T(x) $$ is fully reducible (splits completely into linear factors)? Is it true that for any $B$ there exist $H$ and $T$ such that $\S$ is fully reducible? What is the set of all those polynomials $B$ for which $H$ and $T$ can be found so that $\S$ is fully reducible? Is there any reasonably strong condition that $B$ must satisfy in order for such $H$ and $T$ to exist?

$\newcommand\S{\mathcal S}$ Let $p$ be a prime, and suppose that $0.9(p-1)<q<p-1$. Suppose, furthermore, that $H,B,T\in\mathbb F_p[x]$ satisfy $\max\{\deg H,\deg T\}\le q-1$ and $\deg B\le p-1-q$. What can be said about $B$ given that the polynomial $$ \S(x) := x^p H(x) + x^q B(x) + T(x) $$ is fully reducible (splits completely into linear factors)? Is it true that for any $B$ there exist $H$ and $T$ such that $\S$ is fully reducible? What is the set of all those polynomials $B$ for which $H$ and $T$ can be found so that $\S$ is fully reducible? Is there any reasonably strong condition that $B$ must satisfy in order for such $H$ and $T$ to exist?

As an example, if $B$ is the zero polynomial, then one can take $H(x)=1$ and $T(x)=-2x^{\frac{p+1}2}+x$ to get $$ \S(x)=x^p-2x^{\frac{p+1}2}+x=(x^{\frac{p-1}2}-1)^2x $$
which, of course, is fully reducible.