Let $I=\langle 240,x^3-x^2\rangle$ be an ideal of $\mathbb Z[x]$, then does $x^3 -x+30\in\sqrt{I}$ hold? The question can basically be re-written as, does there exist $f,g\in\mathbb Z[x]$ and $n\in\mathbb N$ such that $240f(x) +(x^3-x^2)g(x)=(x^3-x+30)^n$?
I firstly tried to compare constants and the coefficients of $x$ on the left and right hand side but couldn’t reach a contradiction. I think there are no such polynomials or $n$ since it would be difficult to find them explicitly, but not sure if there are some quick tricks or properties of radical ideals that are able to help answer the question
 A: Factoring is your friend. The two parts $x^3-x$ and $30$ of your polynomial each look similar in form to one of the two generators of $I$. Factoring the two parts, and factoring the two generators, and comparing, we find that $30\in\sqrt I$ as $30^4=3375\cdot240$. And we get $x^3-x\in\sqrt I$, as
$$
(x^3-x)^2=(x-1)(x+1)^2\cdot(x^3-x^2)
$$
Finally, $\sqrt I$ is closed under addition, so $x^3-x+30$ is also contained in $\sqrt I$.
A: Since the lowest power of $30$ that is divisible by $240$ is four, I tried dividing $(x^3-x+30)^4$ by $x^3-x^2$. This gave a remainder of $108000 x^2-108000 x+810000$, which, lo and behold, is divisible by $240$.
A: Let $P$ be a prime ideal containing the ideal $I=(240,x^3-x^2)$.

From $240\in P$ it follows that $P$ contains one of $2,3,5$, hence $30\in P$.

From $x^3-x^2\in P$ it follows that $P$ contains one of $x,x-1$, hence $x^3-x\in P$.

From $30\in P$ and $x^3-x\in P$, we get $x^3-x+30\in P$.

Thus $x^3-x+30$ is an element of every prime ideal $P$ containing $I$.

Therefore $x^3-x+30\in\sqrt{I}$. 
