6.352 Solve the equations $x^{4}-x^{3}-22 x^{2}+16 x+96=0$ and $x^{3}-2 x^{2}-3 x+10=0$, given that they have a common root.
Solution. Let $x_{0}$ be a common root,

$$
f(x)=x^{4}-x^{3}-22 x^{2}+16 x+96, g(x)=x^{3}-2 x^{2}-3 x+10
$$

- Represent $f(x)$ in the form (3). Then $r_{1}(x)=-17 x^{2}+9 x+86$.

$$
\text { Further, } \eta(x)=0 \Leftrightarrow\left[\begin{array}{l}
x_{0}=-2 \\
x_{0}=\frac{43}{17}
\end{array} . \text { By checking, we find that } x_{0}=-2\right. \text {, }
$$

from which $f(x)=(x+2)(x-3)(x-4)(x+4), g(x)=(x+2)\left(x^{2}-4 x+5\right)$.

Answer: roots of the first equation: $x_{1}=-2, x_{2}=3, x_{3,4}= \pm 4$;

root of the second equation: $x=-2$.