5. Given the equation in $x$
$$
x^{4}-6 x^{3}+a x^{2}+6 x+b=0
$$

the left side can be divided by $x-1$, and the remainder when divided by $x+2$ is 72. Then all the solutions of this equation (in ascending order) are $\qquad$
5. $-1,1,2,4$.

From the given information, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
1-6+a+6+b=0, \\
(-2)^{4}-6(-2)^{3}+a(-2)^{2}+6(-2)+b=72
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a=7, \\
b=-8 .
\end{array}\right.
\end{array}
$$

Thus, the original equation is
$$
x^{4}-6 x^{3}+7 x^{2}+6 x-8=0 \text {, }
$$

which is $(x+1)(x-1)(x-2)(x-4)=0$.
The solutions to this equation (in ascending order) are $-1,1,2,4$.