Product rule and polynomial

Question

I don't understand the answer, why does $p'(z)$ equal to that equation? And why does that equation dividing by p(z) yield the final answer?

Answer 1

Writing $p(z) = (z-a_1) \cdots (z-a_n)$, you simply find $p'(z) = \frac{d}{dz} p(z)$. If you use the product rule ($(fg)' = f'g+fg'$) on each linear factor, you arrive at the equation.

Dividing by $p(z)$ to get the other relation is just a consequence of the rules for division. Just try it and see if it works out.

Answer 2

The derivative of $z-a$ is just $1$. When you have a polynomial like that, the product rule tells you that you will have the sum of the products of all but one of the monomials. For instance, the derivative of $(x-1)(x-2)(x-3)$ is $(x-1)(x-2)(1)+(1)(x-2)(x-3)+(x-1)(1)(x-3)$. When you divide that by the original polynomial, you get a fraction that can be split up as shown in the sum formula. I would also like to point out that if you use Logarithmic Differentiation, you will get the same result.