\(A=\dfrac{3x^3-4x^2+x-1}{x-4}=\dfrac{3x^2\left(x-4\right)+8x\left(x-4\right)+33\left(x-4\right)+131}{x-4}=\dfrac{\left(x-4\right)\left(3x^2+8x+33\right)+131}{x-4}=3x^2+8x+33+\dfrac{131}{x-4}\in Z\)
\(\Rightarrow\left(x-4\right)\inƯ\left(131\right)=\left\{-131;-1;1;131\right\}\)
\(\Rightarrow x\in\left\{-127;3;5;135\right\}\)