If $R$ is Armendariz then $R[x]$ is Armendariz.

Question

A ring $R$ is said to be Armendariz if whenever $$f(x),g(x)\in R[x]$$ satisfy $f(x)g(x)=0$ then $a_ib_j=0,\forall 0\leq i \leq n, 0\leq j \leq m $, where $f(x)=\sum_{i=0}^{i=n}a_ix^i$ and $g(x)=\sum_{j=0}^{j=m}b_jx^j$. Now this result is proved by D.D Anderson and Victor Camillo in their paper "Armendariz rings and Gaussian rings", but I didn't understand fully. They proved like this way: Take $f(T),g(T)\in R[x][T]$ such that $f(T)g(T)=0$ where $f(T)=f_0+f_1T+...f_nT^n$ and $g(T)=g_0+g_1T+...+g_mT^m$, then he took that suppose $k=degf_0+...+degf_n+degg_0+...+degg_n$ and then write that $f(x^k)g(x^k)=0$, I don't understand this step. Please clarify.