Show that the ideal of all polynomials of degree at least 5 in $\mathbb Q[x]$ is not prime 
Let $I$ be the subset of $\mathbb{Q}[x]$ that consists of all the
  polynomials whose first five terms are 0.

I've proven that $I$ is an ideal (any polynomial multiplied by a polynomial in $I$ must be at least degree 5), but I'm unsure to how to prove that it is not a prime ideal.  My intuition says that its not, because we can't use $(1)$ or $(x)$ as generators.  
I know that $I$ is a prime ideal $\iff$ $R/I$ is an integral domain.
Again, I'm a little confused on how represent $\mathbb{Q}[x]/I$    

Hint $\ $ For any prime ideal $\rm\,P\!:\,\ x^n\in P\:\Rightarrow\:x\in P.\:$ Thus $\rm\ x^5 \in I\,$ but $\rm\ x\not\in I\ \Rightarrow\  I\,$ is not prime.
Equivalently, $\rm\, R\ mod\ P\,$ has a nilpotent ($\Rightarrow$ zero-divisor): $\rm\, x^5\equiv 0,\ x\not\equiv 0,\,$ so it is not a domain.
Remark $\ $ Generally prime ideals can be generated by irreducible elements (in any domain where nonunits factor into irreducibles), since one can replace any reducible generator by some factor, then iterate till all generators are irreducible. In particular, in UFDs, where irreducibles are prime, prime ideals can be generated by prime elements. This property characterizes UFDs. Well-known is Kaplansky's case: a domain is a UFD iff every prime ideal $\!\ne\! 0$ contains a prime $\!\ne\! 0.$
