Prove or disprove that the ideal $\Big((X+1)^2,(2X+1)(X^2-2)\Big)$ is a principal ideal in $\mathbb{Z}[X]$
Assume that the ideal $\Big((X+1)^2,(2X+1)(X^2-2)\Big)$ is principal. Then exist generator $f(X)\in \mathbb Z[X]$ such that: $$\Big((X+1)^2,(2X+1)(X^2-2)\Big)=f(X)\mathbb Z[X]$$ However I don't have idea how to (dis)prove that $f$ exist.
Division with remainder gives
$2x^3+x^2-4x-2 = (2x-3)(x^2+2x+1) + 1$.
Thus $1$ lies in the ideal and so the ideal equals $\Bbb Z[x]$.
$\begin{align}{\bf Hint}\ \ I =&\ ((x\!+\!1)^2,\ f(x))\\[.2em] = &\ ((x\!+\!1)^2,\ f(x)\bmod (x+1)^2)\ \ \ {\rm as\ in\ Euclidean\ Algorithm}\\[.2em] = &\ ((x\!+\!1)^2,\ \color{#c00}{f(-1)} + \color{#0a0}{f'(-1)} (x\!+\!1))\ \ \ {\rm by\ Taylor\ expansion} \end{align}$
Your $\,f(x)\,$ has $\,\color{#c00}{f(-1) = 1}\,$ and $\,\color{#0a0}{f'(-1) = 0}\,$ so $\,\color{#c00}{1}\in I\Rightarrow\, I = (\color{#c00}{1})$
Remark $\ $ i.e. $f\equiv 1\pmod{(x\!+\!1)^2}\,$ by double root test shows $-1$ is a double root of $\,f-1$
