5. Let $(x+\alpha)^{4}=x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}$. If $a_{1}$ $+a_{2}+a_{3}=64$, then $\alpha$ equals ( ).
(A) -4
(B) 4
(C) -2
(D) 2
5. D.

Let $x=1$, then
$$
(1+\alpha)^{4}=1+a_{1}+a_{2}+a_{4}+a_{4}=65+a_{4} \text {. }
$$

Let $x=0$, then $\alpha^{4}=a_{4}$.
$$
\therefore(1+\alpha)^{4}-\alpha^{4}=65 \text {. }
$$

This leads to $2 \alpha+3 \alpha^{2}+2 \alpha^{3}-32=0$,
which is $(\alpha-2)\left(2 \alpha^{2}+7 \alpha+16\right)=0$.
ifij $2 \alpha^{2}+7 \alpha+16 \neq 0$,
$$
\therefore \alpha=2 \text {. }
$$