Am looking to prove following conjecture: Am looking to prove that the tangent to the cubic $y = x^3 + 7x^2 + 4x + 28$ at any point midway between any two roots will itself cut the x-axis at the remaining root. The cubic factorizes as $y = (x+2i)(x-2i)(x+7)$. And here's my problem: complex roots. How can I solve the above?
 A: The claim is true for any cubic, whether the roots are real or not. Look at this:

The above is Mathematica output. Here is what it does:
We begin with an arbitrary cubic polynomial $p(x):=(x-a)(x-b)(x-c)$ with given real or complex zeros $a$, $b$, $c$ and put
$$y_0:=p\left({a+b\over2}\right)=-{1\over8}(a-b)^2(a+b-2c),\quad y_1:=p'\left({a+b\over2}\right)=-{1\over4}(a-b)^2\ .$$
In the case where the roots are real the equation
$$y:=q(x):=y_0 + y_1 \left(x-{a+b\over 2}\right)$$
describes the tangent to the graph of $p$ at $x_0:={a+b\over2}$. The last command ("Solve") in the above Mathematica dialog computes the zero of the linear function $q$, and sure enough it found $x=c$.
A: The only way I can see you can try in this case is taking all the roots as points in $\,\Bbb C\cong\Bbb R^2\,$ , so that the roots are $\,(-7,0)\,,\,(0,2)\,,\,(0,-2)\,$ , but then you have a new problem (or perhaps you only need a good definition): is the midpoint of the line segments between any two of these points a point on the function's graph? No, it's not. For example, the midpoint between points $\,2\,,\,3\,$ is the origin $\,(0,0)\,$ , which is not part of the graph, or the midpoint between points $\,1\,,\,2\,$ , which is $\,\displaystyle{\left(-\frac72\,,\,1\right)}\;$, and it is also not on the graph.
Thus, the conjecture is false under the above assumption. If you have other ones write them down.
A: Your conjecture is already proved correct by Christian Blatter.
In case of complex roots also it is quite correct, presenting no problem either theoretically or in its geometrical/graphical presentation. Only difference is that you need to draw the tangent line at average value of the complex conjugate roots which is nothing but the real part, in the case you gave it is zero. I attach a sketch to illustrate this aspect. 
The slope of tangent is differential coefficient of
$ x^3+ 7 x^2 + 4 x + 28 $ at $ x = (- 2 i + -2 i)/2, $ i.e.,$ x = 0, $ 
is clearly 28/7 = 4 of the blue line tangent.


EDIT2: Another example of a cubic with complex and real roots:
$ f(x)=(x - 2 + i) (x - 2 -i)(x+3) = 0 $
Summing up,
Tangent at real part of complex root or
average of real roots
always cuts f(x) at its real root. 
EDIT3:
The following may be useful/interesting formulas relating to cubic equations:
When roots are real: $ f(x)= (x-a)(x-b)(x-c)$, between roots a and b,
 i.e.,at mid-point x =(a+b)/2, the slope of real root-tangent extremum is 
 $(b-a)^2/4 $, extremum value = slope.$ ((a+b)/2-c) $. 
When one root is real and two complex: $ f(x)= (x-a-i b )(x-a + i b)(x-c) $,  for roots @ $ x=a  \pm i b $, real root that should be taken is $ x = a$ only,  the slope of real root-tangent extremum is simply $ b^2 $, extremum value = $ b^2  (a-c). $ 
