Prove that if P(x) is a quartic polynomial, then there can only be at most one line that is tangent to it at two points. Prove that if P(x) is a quartic polynomial, then there can only be at most one line that is tangent to it at two points.
From what I have seen, the only case in which I have found a tangent line at 2 points is when the tangent line is horizontal. Is there any other way to approach it?
 A: Suppose $P(x)$ is quartic and $T(x)=Ax+B$ is a line that is tangent to $P(x)$ at $\xi_1,\xi_2$. Then you can factor $P-T$ as
$$P(x)-T(x) = (x-\xi_1)^2(x-\xi_2)^2$$
Then $$P''(x)= 2((x-\xi_1)^2+4(x-\xi_1)(x-\xi_2)+ (x-\xi_2)^2))$$
So that $P''(\xi_1) = P''(\xi_2)= 2(\xi_1-\xi_2)^2$. This determines $\xi_1,\xi_2$ uniquely, since they must be symmetric points with respect to the vertex of the quadratic $P''$. 
In the example given in the comments, $P(x)=  x^4-2x^2+x+1$ so $P''(x) = 12x^2 -4$ and $\xi_1=-\xi_2$. Hence 
$$12\xi_1^2- 4= 2(\xi_1+\xi_1)^2$$
$$\xi_1^2 = 1$$
whence $\xi_1 = -\xi_2 = 1$. In this case,
$$P(x)-T(x) = (x-1)^2(x+1)^2$$
A: Line $y=ax + b$ is tangent to polynomial $y = P(x)$ at $x_1$ iff $P(x_1)=ax_1+b$ and $P'(x_1)=a$ which is equivalent to $Q(x)=P(x)-ax-b$ having a double root $Q(x_1)=Q'(x_1)=0$ at $x_1$.
Given $P(x)$ is a quartic then $Q(x)$ is a quartic as well, thus it can have at most $4$ real roots (counting multiplicities). Therefore $Q(x)$ can not have more than $2$ double roots, so there are at most $2$ points of tangency between the quartic and any line.
