What are the prime ideals of $K[x]/\langle x^n\rangle$? What are the prime ideals of $K[x]/\langle x^n\rangle$, where $K$ is a field?

I have tried it like this: suppose its prime ideal be $P/\langle x^n\rangle $. Then $x^n$ belongs to $P$, which implies that $x$ is in $P$ since $P$ is a prime ideal of $K[x]$. Now i am confused further how to solve. Is $P= K$?
Consider the general case: $K[x]/\langle f(x)\rangle$.

There is a bijection between the ideals in $K[x]/\langle f(x)\rangle$ and the ideals in $K[x]$ that contain $\langle f(x)\rangle$. This correspondence respects prime ideals.

Since $K[x]$ is a PID, the ideals that contain  $\langle f(x)\rangle$ are exactly those of the form $\langle p(x)\rangle$, where $p$ is an irreducible divisor of $f$.

Therefore, the prime ideals $K[x]/\langle x^n\rangle$ correspond to the prime divisors of $x^n$, which is just $x$.
Recall that $Spec A/\alpha =\{ p\in Spec A \mid \alpha \subset p\}$.So we can get $Spec k[x]/&lt;x^n&gt; =Spec k[x]/&lt;x&gt;$.we know $&lt;x&gt;$ is a maximal ideal of $k[x]$,so $Spec k[x]/&lt;x^n&gt;$ contains only one element.