Let $ I = \langle 5x^2 \rangle$ be an ideal of $\mathbb{Z}[x]$
Let $J_{x+1} = {\{f(x) \in \mathbb{Z}[x] : f(x)\cdot(x+1) \in I}\}$. How do I find out what this ideal is?
I assume for a polynomial $g(x)$ to be in $I$ it would have to be divisible by $5x^2$, so if we looked at $J_x$, then $f(x)$ would have to be divisible by $5x$, and so $J_x = \langle 5x \rangle$.
I can’t seem to figure out what $J_{x+1}$ is though? My intution tells me it is $I$ but I’m not sure.
From the comments, for $g \in \Bbb{Z}[x]$, you are defining:
$$ J_g = \{ f \in \Bbb{Z}[x] : fg \in I\} $$
where $I$ is the ideal $\langle 5x^2 \rangle$ and you want to know what $J_g$ is when $g = x + 1$. (Here I am writing $f$ where you write $f(x)$ as is standard in this context: we don't need a continual reminder that $f$ is a polynomial in the indeterminate $x$.)
Now, $f \cdot (x+1) \in I$ iff there is $h \in \Bbb{Z}[x]$ such that:
$$ f \cdot (x + 1) = h \cdot 5x^2 $$
but $x + 1$ is irreducible and it is not a factor of $5x^2$ (as you can check by polynomial division), so if the above equation holds, $x + 1$ is a factor of $h$ (using the fact that $\Bbb{Z}[x]$ is a unique factorisation domain). I.e., we have $$ h = h_0 \cdot (x + 1) $$ for some $h_0 \in \Bbb{Z}[x]$ and then we have $$ f = h_0 \cdot 5x^2 $$ implying that $f \in I$.