The roots of a polynomial and of its gradient function
An example
The roots of the cubic
f  x   x3  6 x 2  9 x  2
  x  2   x 2  4 x  1
are x  2, x  2  3 and x  2  3 .
2
y
x
1
2
3
4
−2
The mean average of the roots of f  x   0 is

 
2 2 3  2 3
3
Differentiating the function:
f   x   3 x 2  12 x  9  3  x  1 x  3 .
This has roots x  1 and x  3
The mean average of the roots of f   x   0 is
1 3
2
2
These two means are the same!
Is this a coincidence or do all cubics behave this way?
What about other polynomials?
2
A look at the general case
Take a polynomial of order n . This will have n roots, some of which may be complex.
Assuming the coefficient of x n is 1, the polynomial can be written in the form
f  x    x  a1  x  a2  ...  x  an  .
This can be expanded:
f  x   x n  Ax n 1  ...  Bx  C , where A  a1  a2  ...  an .
This implies that the mean average of the n roots, whether real or complex, of the
A
equation f  x   0 is .
n
Differentiating f  x  gives
f   x   nx n 1  A  n  1 x n  2  ...  B .
Therefore, the sum of the n  1 roots, whether real or complex, of the equation
A  n  1
b
f   x   0 must be
(using the familiar  rule) and so the mean of these
n
a
1 A  n  1 A
roots is
 .
n
 n  1 n
For all polynomials the mean of the roots is equal to
the mean of the roots of the derived function.