Why can't a prime number p be the base of a maximal ideal of R[x]? R[x] is a polynomial ring. Come to think about it, if you have R(x).p, Q(x).p, etc, all these can be members of the aforementioned ideal. 
The textbook I'm referring to says only linear polynomials of the form (x-a) can be the base for a maximal ideal of R[x]. 
 A: Answering the question in the title:
Let $p\in R$ be a prime element, so that $R/(p)$ is a domain. Then consider $I=(p)R[X]$, the ideal generated by $p$ in $R[X]$, and note that
$$
R[X]/I\simeq \left(R/(p)\right)[X]
$$
is not a field, hence $I$ isn't maximal in $R[X]$.
A: A maximal ideal in a ring $S$ is defined to be a proper ideal $I$ such that if $J$ is another ideal of $S$ satisfying $I\subseteq J\subseteq S$ then $I=J$ or $J=S$.  Now, if $p$ is a prime element of the ring $S=R[X]$, then $p$ is not a unit, so the polynomial $X$ is not contained in $(p)$ (as if $sp=X$ then $s$ must be a polynomial of the form $rX$ with $rs=1$, which is a contradiction).
Therefore, the ideal $(p,X)=\{\textrm{polynomials whose constant term is a multiple of }p\}$ contains $(p)$, and is contained in $R[X]$, but it is not equal to either of them.  So $(p)$ is not maximal.  
A: I assume calling $p$ a "prime number" implies that it is an element of $R$. Then it cannot generate a maximal ideal because $R[X]/(p)\cong (R/(p))[X]$ is either the trivial ring (if $p$ is invertible) or a polynomial ring; in either case it is not a field.
A: Hint $\ \rm\:n\in R,\ (n)\:$ maximal in $\rm\:R[x]\:\Rightarrow\: (n) = (1)$
Proof $\ $ If $\rm\  x\in (n)\:$ then $\rm\:x = n f\,\stackrel{x=1}{\Rightarrow} 1 = n\,f(1)\:\Rightarrow\: (n) = (1).\:$  Else $\rm\: x\not\in (n),\:$ so $\rm\,(n)\,$ maximal $\rm\:\Rightarrow\:(x,n) = (1),\:$ so $\rm\:xf+ng = 1\stackrel{x=0}{\Rightarrow}n g(0) = 1\:\Rightarrow\:(n)=(1). $
