Intersections of Polynomials

Given a cubic with 3 distinct real roots f(x) = (x-a)(x-b)(x-c), 
pick any 2 of those roots, say, a and b, and let v = (a + b)/2.  
Then the line tangent to the graph of f(x) at the point (v,f(v)) 
contains the point (c,0).  The proof of this is elementary, but
it's interesting to explore it, to see WHY it's true.

The algebra isn't really too messy if we remember that the coefficient 
of x^(n-1) of a monic nth degree polynomial is the (negative) sum of 
the roots.  If our tangent line is L(x) the cubic h(x) = f(x) - L(x) 
has a double root at v, so its roots are [v,v,q] whereas the roots of 
f(x) are [a,b,c].  Since h(x) and f(x) have the same coefficient for 
x^2 we have (a+b+c) = (v+v+q).  Since we stipulated that v=(a+b)/2
it follows that a+b = v+v and so q=c, as expected.

To give a slightly more "visual" geometrical explanation, notice that 
the proposition is actually a special case of the fact that for ANY 
real number q the line through the two points 

    [(m+q), f(m+q)]      [(m-q), f(m-q)]      where  m=(a+b)/2

passes through the x axis at c.  (The particular case we started with 
corresponds to q = 0.)  To visualize why this is so, notice that 
the quadratic polynomial g(x)=(x-a)(x-b) is just a parabola whose
directrix is the vertical line bisecting the interval [a,b].  In
other words, it is the vertical line at x = (a+b)/2.  Since the
parabola is symmetrical about this line, we have g(m+q)=g(m-q) for
all q.  Thus, the two points noted above are

      [ (m+q), (G)(m+q-c) ]     and   [ (m-q), (G)(m-q-c) ]

where G = g(m+q) = g(m-q).  Where will the line through these two
points cross the x axis?  The value of G is just a scale factor
that doesn't affect where the line crosses the x axis, so we just
need to picture two points of the form

              [ x1, (x1-c) ]     [ x2, (x2-c) ]

These are as shown (crudely) in the drawing below

                |              /
                |            / |  /
              y2|          /   o/
                |        / |  /|
              y1|      /   o/  |
                |    /    /|   |
                |  /    /  |   |
                |/____/____|___|______
                     c    x1   x2

Since the two points are a distance c below the diagonal, they 
obviously strike the x axis at c.  In the special case q=0 the
two points merge into one, and the direction of the tangent line
is the limit of the chords as the points approach each other.

The generalization to higher degrees is most easily made for the
converse fact, i.e., if a straight line passing through the point 
(c,0) strikes the cubic (x-a)(x-b)(x-c) at two points, then the x 
coordinates of those two points are equidistant from (a+b)/2.
Similarly, a vertical parabola passing through the points (c,0) 
and (d,0) strikes the quartic (x-a)(x-b)(x-c)(x-d) at two points 
whose x coordinates are equidistant from (a+b)/2.

In general, if the polynomial

    f(x)  =  (x - r_1)(x - r_2)(x - r_3)...(x - r_k)

strikes the polynomial  h(x) = (x-a)(x-b)f(x)  at two points, then 
the x coordinates of those two points are equidistant from (a+b)/2.