A common tangent line 
The graph of $f(x)=x^4+4x^3-16x^2+6x-5$ has a common tangent line at $x=p$ and $x=q$. Compute the product $pq$.

So what I did is I took the derivative and found out that $p^2+3p+q^2+3q+pq=0$. However when I tried to factorize it I didn't find out an obvious solution. Can someone hint me what to do next? Thanks in advance

Consider the quartic polynomial $P(x)=f(x)-ax$ and let $t:=x+b$. Find $a$, $b$ such that $t\to P(t-b)$ is an even function by requiring that the coefficients of $t$ and $t^3$ are zero. It turns out that $b=1$ and $a=46$.
Hence $P(t-1)=t^4-22t^2+16$ whose derivative is $P'(t-1)=4t(t^2-11)$. Therefore $P$ has two minimum points at $t_1=-\sqrt{11}$ and $t_2=\sqrt{11}$ with a common horizontal tangent. Assuming $p<q$, it follows that
$$p=t_1-b=-\sqrt{11}-1\quad\mbox{ and }\quad q=t_2-b=\sqrt{11}-1\implies pq=-10.$$
