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
 A: Hint:
If the line $y=mx+n$ is tangent of the curve $y=f(x)=x^4+4x^3-16x^2+6x-5$ at the points $P=(p,f(p))$ and $Q=(q,f(q))$, this means that the system
$$
\begin{cases}
y=x^4+4x^3-16x^2+6x-5\\
y=mx+n
\end{cases}
$$
has the two double solutions $x=p$ and $x=q$. And this means that the polynomial $x^4+4x^3-16x^2+6x-5-mx-n$ factorize as $(x-p)^2(x-q)^2$.
Using the identity principle for polynomial, from 
$$
x^4+4x^3-16x^2+6x-5-mx-n=(x-p)^2(x-q)^2
$$
you can find $p$ and $q$.
Note: this is a solution that does not use the derivatives (only algebra, no calculus), but you can solve the problem also noting that $y=f(x)$ has a''bi-tangent''at $x=p$ and $x=q$ iff:
$$
\frac{f(p)-f(q)}{p-q}=f'(p)=f'(q)=m
$$
where $m$ is the slope of the bi-tangent line.
A: Let the equation of the common tangent be $y=mx+b$.
Solving simultaneously the equations of the common tangent and $f(x)$, we get
$x^4+4x^3-16x^2+(6-m)x-(b+5)=0$
This equation will have two double roots (since a line is touching a curve twice), so let the roots be $p,\,p,\,q,\,q$.
Can you take it from here?
A: 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
$$p=t_1-b=-\sqrt{11}-1\quad\mbox{ and }\quad q=t_2-b=\sqrt{11}-1\implies pq=-10.$$
A: Since the specific gradient is not given, I think it would be appropriate to leave it in a very general form. If
$$
p^{2} + 3p + q^{2} + 3q + pq = 0
$$
were correct (which we have verified is not true), then
$$
pq = -\left(p^{2} + q^{2} + 3p + 3q\right).
$$
Out of personal preference, I would rewrite
$$
p^{2} + pq + q^{2} + 3p + 3q = (p + q)^{2} - pq + 3(p + q) = 0
$$
so that
$$
pq = (p + q)^{2} + 3(p + q).
$$
A: The tangent line equation at $x=p$ and $x=q$ is:
$$y=p^4+4p^3-16p^2+6p-5+(4p^3+12p^2-32p+6)(x-p);\\
y=q^4+4q^3-16q^2+6q-5+(4q^3+12q^2-32q+6)(x-q).\\$$
By equating the coefficients, simplifying and denoting $a=p+q, b=pq$ we get:
$$\begin{cases}b=a^2+3a-8\\
3a^3-(6a+8)b+8a^2-16a=0\end{cases}$$
Solving the system we get:
$$a=-2,b=-10=pq.$$
