Is SL(n,Z[x]) generated by transvections? Is $\mathrm{SL}(n,\mathbb{Z}[x])$ equal to $E(n,\mathbb{Z}[x])$, the subgroup generated by transvections?  
 A: First, for $n=2$, the result is certainly false by the counterexample of Cohn quoted in Oblomov's comment.
For higher $n$, start with the fact that the Bass Stable Rank of a commutative ring is at most 1 more than the Krull dimension.  (I'm sure you can find this hidden in Chapter V of Bass's book on K-theory, though extracting it might require some work to get familiar with his notation).  Therefore ${\mathbb Z}[X]$ has stable rank at most 3.  In fact, it's almost surely exactly 3.  (There is some discussion here about whether this stable rank might in fact be 2, but there's something close to a proof there that this is not the case.)
Now from Bass, Chapter V, 3.3, it follows that for all $n\ge 4$, we have
$$GL_n({\bf Z}[x])=GL_3({\mathbb Z}[X])E_n({\mathbb Z}[X])$$
In other words, for any matrix of size $4x4$ or greater, you can, by applying elementary operations, reduce to a matrix of the form 
$$\pmatrix{A&0\cr 0&I\cr}$$
where $A$ is 3 by 3 and $I$ is the $(n-3)\times (n-3)$ identity matrix.
It's not immediately clear to me how much better you can do, but I bet either that Wilberd van der Kallen (who shows up here occasionally) or Vaserstein (who I think does not) could answer this in his sleep.  You might want to email one or both of them.
