Why does substitution x+1 preserve irreducibility of polynomials?

[nasshi]

This is for clarification of a method.

Dummit & Foote, pg 310, Example (3).

$f(x)=x^{4}+1$ is converted into $g(x)=f(x+1)$ in order to use Einsenstein's Criterion for irreducibility. The example states "It follows that $f(x)$ must also be irreducible, since any factorization of $f(x)$ would provide a factorization of $g(x)$ (just replace $x$ by $x+1$ in each of the factors)."

My question is, "In each of the factors of what?". $f(x)$ if it were factorable? In $g(x)$ since $f(x)$ was theoretically factorable by their explanation?

Please provide a more detailed explanation if possible. An example of this technique when a polynomial is reducible would be great. I was unable to create one since the wording has confused me.

 

[voko]

If $f(z)$ is reducible, then $f(z) = p(z)q(z)$ These are the factors the book is talking about.

 

[nasshi]

So as an example, if defining $g(x)=f(x^{2}+45x-2)$ and Eisenstein's criterion showed that $g(x)$ is irreducible, then $f(x)$ is irreducible? Or can I only use linear factors such as $g(x)=f(x-2)$?

 

[voko]

I think you could prove a more general statement, but linear polynomials are obvious.