# Polynomial roots property [duplicate]

Let $$P$$ be a $$n$$-degree real polynomial having $$n$$ simple roots $$x_1, x_2, \cdots,x_n$$.

The problem asks to prove that

$$\displaystyle\sum_{k=1}^n \frac{P''(x_k)}{P'(x_k)} =0$$

I tried writing $$P$$ as the product of $$x−x_i$$ and differentiating it twice but the calculations are too hard to carry out. Thanks in advance.

## marked as duplicate by Martin R, Song, Carl Schildkraut, Cesareo, Parcly TaxelMar 1 at 1:14

• I tried writing $P$ as the product of $x-x_i$ and differentiating it twice but the calculations are too hard to carry out – ahmed Feb 28 at 18:32
• @ahmed Can you differentiate it once? Notice how simple $P’$ looks when you evaluate it at an $x_i$. Can you see a way to simplify $P’’$ also? (it will probably still have about $n$ terms but it’s manageable) – Erick Wong Feb 28 at 18:42
Let $$P(x)=(x-1)(x+1)(x-2)$$ Then $$P'(x)=1.(x+1).(x-2)+(x-1).1.(x-2)+(x-1).(x+1).1$$ and $$P''(x)=1.1.(x-2)+1.(x+1).1+1.1.(x-2)+(x-1).1.1+1.(x+1).1+(x-1).1.1$$ $$P''(x)=2 [(x-1)+(x+1)+(x-2)]$$ $$P''(1)=2[2-1] =2 : P''(-1)=2[-2-3]=-10 : P''(2)=2[1+3]=8$$ And $$2-10+8=0$$ So, now it's about generalising this process...