Is there a polynomial $f(x) \in \mathbb{Q}[x]$ of degree $d$, such that on changing any $O(\sqrt{d})$ of its coefficients, it still remains irreducible?
By changing, I mean I have no control over what those changed $O(\sqrt{d})$ coefficients would be. Now of course, if the constant term becomes zero after change, then we will have $x$ as a factor, and hence reducible. Say, I am okay with such changes. In other words, is there an explicit polynomial $f(x)$ such that after change, it becomes $x^i.f^{\prime}(x)$, where $f^{\prime}(x)$ is irreducible and $i \geq 0$.
I was thinking to start with a polynomial with different prime numbers as coefficients that satisfies Eisenstein's or some irreducibility criterion, so that it becomes easier to argue irreducibility after change also, by looking at remaining $d-O(\sqrt{d})$ coefficients. But, I was unable to find any irreducibility criterion that is compatible with change of coefficients.
Any positive or negative answer to the above question is appreciated, in fact, for $f \in \mathbb{F}[x]$, where $\mathbb{F}$ is any field. I also don't mind if there is a result which is not precisely $O(\sqrt{d})$ coefficients but is anything of this flavor that after perturbation of some of its coefficients, the polynomial remains irreducible.
Let $d$ be an integer with $d\ge 2$, and let $f\in \mathbb{Z}[x]$ be irreducible with $\deg(f)=d$.
Write $f={\displaystyle{\sum_{k=0}^d a_kx^k}}$.
Let $b_1=a_1-f(1)$, and let $g$ be the polynomial obtained from $f$ by replacing the term $a_1x$ by the term $b_1x$.
Then $g=f+(b_1x-a_1x)$, hence $g(1)=f(1)+(b_1-a_1)=0$, so $g$ is reducible.
Note that in the transition from $f$ to $g$, only one coefficient was changed.