Tangent line to a Polynomial I was reading up about Polynomial long division yesterday and one of its applications was the ability to find a tangent line to a polynomial without differentiation. 
It stated let $ P(x)$ be a polynomial, to find the tangent line to $P(x)$ at point $x=k$ divide $P(x)$ by $(x-k)^2$ and the remainder $R(x)$ will be the equation of the tangent line at point $ x = k$.
How is this even possible? I thought that the only way to find the gradient/slope of a curve was through differentiation.
 A: In math, it is pretty rare and unnatural to have only one way to do something.
Really, what is a remainder? We have $P(x)=(x-k)^2Q(x)+R(x)$, where $R(x)$ is a linear function. Now, what is the derivative of $P$ at $k$? Can you find it? See: $P'(x)=2(x-k)Q(x)+(x-k)^2Q'(x)+R'(x)$, so when we plug $x=k$, the first two terms vanish and leave us with $R'(k)$. Also, quite obviously, $P(k)=R(k)$. So $R(x)$ is a linear function which has the same value and the same slope as $P(x)$ at $x=k$. Sounds mighty like the definition of a tangent line, doesn't it?
A: There is another way of seeing this. Let $I$ be the ideal generated by $(x-k)^2$. This ideal consists of all polynomials that have a root $k$ with multiplicity $2$, another way of saying that the are tangent to the $x$-axis at the point $x = k$.Any representative of a coset of $I$ has an identical tangent line at the point $x = k$.
A: Here's an intuitive non-calculus based answer.


*

*First after the division we have $ f(x) = (x-k)^2Q(x) + R(x)$ where R(x) is a linear equation.  So at x = r, the quotient term zeroes out and f(k) = R(k).  In other words there is an intersection between the curve and line.

*If the line R(x) is tangent at that intersection then f(x) - R(x)  must be either negative or positive for a small delta around x and on both sides. If its a regular intersection then the difference will change signs on the left and right.

*Because we chose to divide by $ (x-k)^2$ after adjusting by a small delta  $ (x - k \pm \Delta)^2 >= 0$. As long as k is not a root of Q(x) its zeros are far enough away from the delta we chose that the delta won't cross them and its sign cannot shift either.  So the necessary condition is met for tangency.


Note: the procedure breaks down when Q(x) does have additional roots of k but then R(x) = 0.
