Average of roots equals average of roots of derivative

As stated on the previous page, for the polynomial a_n zⁿ + a_{n − 1} z^{n − 1} + ··· + a₁ z + a₀, the sum of the roots is given by −a_{n − 1}/a_n. Therefore the average of the roots is given by −a_{n − 1}/(na_n). Now the derivative of this polynomial is na_n z^{n − 1} + (n − 1)a_{n − 1} z^{n − 2} + ··· + a₁. Arguing as before, the average of the roots of the derivative is

This shows that the average of the roots of the derivative is equal to the average of the roots of the original polynomial.