Two points on curve that have common tangent line Find the two points on the curve $y=x^4-2x^2-x$ that have a common tangent line.
My solution: Suppose that these two point are $(p,f(p))$ and $(q,f(q))$ providing that $p \neq q$. Since they have a common tangent line then: $y'(p)=y'(q),$ i.e. $4p^3-4p-1=4q^3-4q-1$ and after cancellation we get: $p^2+pq+q^2=1$.
Tangent lines to curve at points $(p,f(p))$ and $(q,f(q))$ are $y=y(p)+y'(p)(x-p)$ and $y=y(q)+y'(q)(x-q)$, respectively. I have tried to put $x=q$ in the first and $x=p$ in the second equations but my efforts were unsuccesfull.
Can anyone explain me how to tackle that problem?
 A: Tangent:  $ f(p) + f'(p)(x-p) = f(q) + f'(q)(x-q) $  
It follows: 
$(1)\,$ : $\enspace f'(p) = f'(q)\enspace => \enspace p^3-q^3 = p-q $
$(2)\,$ : $\enspace f(p)-pf'(p) = f(q)-qf'(q)\enspace => \enspace 3(p^4-q^4) = 2(p^2-q^2) $
$(1)$ into $(2)$ leads to $\enspace 3(p^4-q^4) = 2(p+q)(p-q) = 2(p+q)(p^3-q^3)$ 
which is equivalent to $\enspace (p-q)^2 (p^2-q^2) = 0\,$ .
Because of the condition $\,p\neq q\,$ we get $\,q=-p\,$ . 
Putting this result into $(1)$ and choosing $p>0$ we get $p=1$ and therefore $q=-1$ .
The points are $\,(1;f(1))=(1;-2)\,$ and $\,(-1;f(-1))=(-1;0)\,$ .
 
A: If $y=mx+n$ is an equation of the tangent to the graph of the  polynomial $f$
then there are real $a$ and polynomial $q$ for which $f(x)-mx-n=(x-a)^2q(x)$. 
Here $a$ it's an abscissa of the touch point. 
Indeed, $f(a)=ma+n$ and $f'(a)=m$.
Thus, $f(x)-mx-n=0$ for $x=a$,
which by Polynomial remainder theorem gives $$f(x)-mx-n=(x-a)h(x)$$ for some polynomial $h$.
Now, $$\left(f(x)-mx-n\right)'=\left((x-a)h(x)\right)'$$ or
$$f'(x)-m=h(x)+(x-a)h'(x).$$
Hence, $$0=f'(a)-m=h(a),$$ which by Polynomial remainder theorem again 
says that there is polynomial $q$,
for which $h(x)=(x-a)q(x).$
Id est, $$f(x)-mx-n=(x-a)^2q(x).$$
Since a degree of $x^4-2x^2-x$ is four,
we need $$x^4-2x^2-x-(mx+n)=(x^2+px+q)^2,$$
which gives the following
$$x^4-2x^2-x=x^4-2x^2+1-x-1=(x^2-1)^2-(x+1).$$
Thus, $y=-x-1$ is an equation of the tangent. 
Now, we got touching points: $(1,-2)$ and $(-1,0)$.
A: Complete the square :
your curve is $y = x^4 - 2x^2 - x = (x^2 - 1)^2 + (-x-1)$. 
So the curve stays above the line $y = -x-1$, and is tangent to it when it touches it, that is when $x^2-1 = 0$, so that's when $x=1$ and $x=-1$ :
The line $y=-x-1$ is tangent to the curve at those two points $(1,-2)$ and $(-1,0)$.
