Help to understand this polynomial trick

Question

I'm trying to understand this answer which I copy here (I didn't ask to the user because he left MSE).

Could someone verify if my answer is correct and help me to understand the highlighted statement, maybe giving me a concrete example?

My attempt:

$\Leftarrow$

If $f$ is reducible, then $f(x)=h_1(x)h_2(x)=g(x+c)$, then $g(x)=h_1(x-c)h_2(x-c)$, so $g(x)$ is reducible.

$\Rightarrow$

If $g(x)$ is reducible, then $g(x)=h_1(x)h_2(x)=f(x-c)$, then $f(x)=h_1(x+c)h_2(x+c)$, so $f(x)$ is reducible.

Thanks in advance

Answer 1

Yes, that's the idea. Basically we are exploiting the fact that the shift map, being an automorphism, faithfully preserves multiplicative structure. Note that you also have to argue that the images of the factors remain nonzero nonunits for the proof to be complete.