jmc

Since $$\begin{array}{rl}(3x-2)(4x+1)-(3x-2)4x+1& =(3x-2)(4x+1-4x)+1\\ & =(3x-2)\cdot 1+1=3x-1,\end{array}$$ when $x=4$ we have the value $3\cdot 4-1=\boxed{11}$.