Suppose $\displaystyle g(x)$ is an irreducible in R. If $\displaystyle f(x)$ is any other polynomial, does that imply that $\displaystyle g(f(x))$ is also irreducible?

I'm particularily interested in the case when $\displaystyle g(x) = x^2 + c$.

Here are some examples I've tried

$\displaystyle g(x) = x^2 + d$

1) $\displaystyle f(x) = c$

$\displaystyle g(f(x)) = c^2 + d$ is a constant

2) $\displaystyle f(x) = x + c$

$\displaystyle g(f(x)) = x^{2} + 2c \cdot x + c^2 + d $

So if I choose (c,d) such that the above has integer roots then I've solved the problem.

So I need $\displaystyle 4c^2 - 4 \cdot (c^2 + d)$ to be a square. But $\displaystyle 4$ divides this, so I just need for $\displaystyle c^2 - c^2 - d = -d$ to be a square.

This is a contradiction to the condition that $\displaystyle g(x) = x^2 + d$ be irreducible because then g(x) would then be a difference of squares.

I stopped after this......Naturally the next f(x) to try would be a quadratic, but I don't know enough about quartic polynomials to know the conditions for when they're irreducible or not.