Problem:
a) Show that if $P(x)$ is a polynomial such that $P(a)=P'(a)=0$ then there exists a polynomial $Q(x)$ such that $[P(x)=(x-a)^2Q(x)$. b) Show that if $P(x)$ is a quartic polynomial then there exists at most one line $\ell$ that is tangent to the graph of $P(x)$ at two places.
I got an answer for the first part meaning that I proved that Q(x) exists. I don't understand how I can prove how there is only one tangent line to the polynomial P(x).
Suppose $l$ and $k$ are lines, tangent to $p$ at $\{a,b\}$ and $\{c,d\}$, respectively.
By part (a) of the problem, \begin{align*} p(x)-k(x)=(x-a)^2(x-b)^2 \\ p(x)-l(x)=(x-c)^2(x-d)^2. \end{align*}
Subtracting one equation from the other we eliminate $p$ and obtain
\begin{equation*} l(x)-k(x) = (x-a)^2(x-b)^2 - (x-c)^2(x-d)^2 \end{equation*}
But the difference of two degree 1 polynomials has degree $\leq$ 1. The degree of the RHS is $\leq 1$ only if $\{a,b\} = \{c,d\}$, i.e., only if $l(x)\equiv k(x)$.