Consider a cubic polynomial, $p(x)=k(x-a)(x-b)(x-c)$ where $k$ is some constant and $a,b,c$ its $3$ roots (not necessarily distinct, not necessarily real). It is very simple to show that if you average two roots of a cubic polynomial and compute the tangent line at their average that it will intersect the cubic polynomial at the remaining root. I tried to generalize this result to other odd degree polynomials and it did not seem to work well.

What is so special about cubic polynomials that allows this to work that isn't true about higher degree odd polynomials?

  • 2
    $\begingroup$ This is not really calculus, but algebra :) $\endgroup$ Commented Aug 18, 2010 at 15:45
  • $\begingroup$ fair enough, I never really know how to tag things :) $\endgroup$
    – WWright
    Commented Aug 18, 2010 at 16:18
  • $\begingroup$ Just out of curiosity, which generalizations have you tried? For example, are you averaging two roots of the higher-degree polynomial, or all the roots except one? $\endgroup$ Commented Aug 18, 2010 at 17:58
  • $\begingroup$ For polynomials of odd degree n, i was averaging n-1 roots and trying to find necessary and sufficient conditions for their average to hit the tangent line, but that clearly wasn't the correct way to generalize. $\endgroup$
    – WWright
    Commented Aug 18, 2010 at 18:10

1 Answer 1


This result does hold for higher-degree polynomials; you just didn't generalize it correctly. The general statement is as follows. Let $p(x) = (x - a) q(x)$ where $q(x)$ is any differentiable function, and let $r$ be such that $q'(r) = 0$. Then the tangent line at $(r, p(r))$ intersects the $x$-axis at $(a, 0)$.

This is a fairly simple computation. Since $p'(x) = (x - a) q'(x) + q(x)$, it follows that $p'(r) = q(r)$, so the tangent line at $(r, p(r))$ has slope $q(r)$ and hence it intersects the $x$-axis at $(r - \frac{p(r)}{q(r)}, 0) = (a, 0)$. In particular, the above result holds for $q(x)$ a polynomial of any degree greater than or equal to $2$.

(Intuitively, when $q'(r) = 0$, the linear approximation to $p(x)$ at $r$ is $(x - a) q(r)$, so of course it has to hit the $x$-axis at $(a, 0)$. In fact, this way of thinking about it tells you that the $r$ such that $q'(r) = 0$ are the only $r$ for which this occurs.)

In the special case that $q(x)$ is quadratic, $r$ happens to be equal to the average of the roots of $q$. For higher-degree polynomials this is no longer the case. Instead, all you know is that the roots of $q'(x)$ interlace between the roots of $q(x)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .