We know that the intersection of all prime ideals of a ring consists of all the nilpotent elements of the ring.
Consider the ring $\Bbb{Z}[x]/(x^2)$. In this ring, the ideal $(2)$ is prime. How do I prove that it contains the nilpotent element $x$?
EDIT: In a hurry, I did not note that $(2)$ is not a prime ideal in this ring. The premise of the question, as of now, is incorrect.
Let $I = (2)$ inside $R = \Bbb Z[x]/(x^2)$. Then $$ R/I\cong \Bbb Z[x]/(x^2, 2)\cong\Bbb F_2[x]/(x^2), $$ which is not an integral domain ($x$ is a zero divisor). That is to say, $(2)$ is not prime in this ring.
In general, let $R$ be a ring, and $\mathfrak{p}\subseteq R$ a prime ideal. An ideal is prime if and only if $ab\in\mathfrak{p}$ implies that $a$ or $b$ is in $\mathfrak{p}$. If $\epsilon\in R$ is nilpotent, then for some $n$, $\epsilon^n = 0$. Because any ideal contains $0$, we have $$ 0 = \epsilon^n = \epsilon\cdot\epsilon^{n-1}\in\mathfrak{p}, $$ so either $\epsilon$ or $\epsilon^{n-1}$ is in $\mathfrak{p}$. Now it follows by induction that $\epsilon\in\mathfrak{p}$. If $(2)$ were prime in your ring, this would prove that $x\in (2)$ abstractly (even though it doesn't explicitly construct $p$ such that $x = 2p$).
The prime ideals of $R/I$ are the prime ideals of $R$ which contain $I$. So in this case, $(2)$ is not a prime ideal of $\Bbb{Z}[x]/(x^2)$ because $(2)\not\ni x^2$ when considered as subsets of $\Bbb{Z}[x]$.
$(x+2)(x-2)=x^2-4\in (2)$ and $x+2,x-2$ are not in $(2)$. Since the coefficients of $(2)(P(x))$ are even where $p(x)\in Z[x]$.