Let P=\ x x^2+(n-1) x-2(n+1)=0\ and Q=\ (n-1) x^2+n x+1=0\ . Then...

Notice that $x=2$ is a root of $P$ because $2^2+(n-1) \cdot 2-2(n+1)=0$ and $x=-1$ is a root of $Q$ because $(n-1) \cdot(-1)^2+n \cdot(-1)+1=0$. Therefore, we see that $P(x)=(x-2)(x+n+1)$ and $Q(x)=(x+1)((n-1) x+1)$ If $|P \cup Q|=3$, then $P(x)$ and $Q(x)$ share a root. Therefore, we have the three equations: \begin{aligned} 2 & =-\left(\frac{1}{n-1}\right) \Rightarrow n=\frac{1}{2}-1=-n-1 \Rightarrow n=0 \\ -\left(\frac{1}{n-1}\right) & =-n-1 \Rightarrow n= \pm \sqrt{2} \end{aligned} Hence, there are total of 7 possible values of $n$.