# Polynomial division and tangent lines

Some time ago I read about a geometric application of polynomial division to finding the equations of tangents to points on a polynomial curve. This is obviously something which is more easily handled with basic differentiation, but still!  I teach polynomial division, factor and remainder theorems about six weeks into my course and mention this application to my students.  Occasionally I get asked to prove it but don’t have time in class.  So here is my sketch of a proof for this claim.

Example: given the curve $f(x) = x^3 + x^2 -4x -6$ We can find the equation of the tangent at the point $$x=2$$ by dividing $$f(x)$$ by $$(x−2)^2$$.

Therefore, the equation of the tangent is $$y=12x−26$$.

The claim is that for any polynomial curve $$y=f(x)$$ then the tangent line at the point where $$x=a$$ can be found by finding the remainder after carrying out the division $$f(x)/(x−a)^2$$.

Here is my attempt at proving the result. First a picture of a function with a tangent to the point $$x=a$$.

First, note that polynomial division results in a quotient and remainder. In this case: $\frac{f(x)}{(x-a)^2} = q(x) + \frac{r(x)}{(x-a)^2}$ which can be easily rearranged to give $f(x) = (x-a)^2 q(x) + r(x)$  A couple of things to note. First $$r(x)$$ must be linear, since it is a degree less than $$(x−a)^2$$, which is quadratic.

Second, at the point $$x=a$$ we find that $f(a) = r(a)$ In other words, $$f$$ and $$r$$ have the same value at $$x=a$$.

The next bit of proof uses some differentiation. Applying the product rule we get: $f^{\prime}(x)=2q(x)(x−a)+q^{\prime}(x)(x−a)2+r^{\prime}(x)$ At the point $$x=a$$ we note that $f^{\prime}(a)=r^{\prime}(a)$ In other words $$f^{\prime}(x)$$ and $$r^{\prime}(x)$$ have the same slope/gradient at $$x=a$$. This is necessary if $$r(x)$$ is a tangent line.

Since $$r(x)$$ is linear, it has the form $r(x)=mx+c$ and we know straight lines are completely specified by knowing their gradient $$f^{\prime}(a)$$  and a point $$(a,f(a))$$ on it.

Therefore $$y=r(x)$$ must be a tangent line to the curve $$y=f(x)$$ at the point $$x=a$$.