4. Given $(x+3)(x-1)^{4}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}$, then $a_{3}+a_{4}+a_{5}=$
$$
\begin{array}{l}
\text { Let } x=1 \Rightarrow a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0 \text {, let } x=0 \Rightarrow a_{0}=3 . \\
a_{1}=1 \cdot(-1)^{4}+3 \cdot C_{4}^{3} \cdot(-1)^{3}=-11, a_{2}=1 \cdot C_{4}^{3} \cdot(-1)^{3}+3 \cdot C_{4}^{2} \cdot(-1)^{2}=14 \text {, }
\end{array}
$$

So $a_{0}+a_{1}+a_{2}=6 \Rightarrow a_{3}+a_{4}+a_{5}=-6$.