An Intriguing Polynomial Problem!

$P(x) = 5(x-x_1)(x-x_2)(x-x_3)\ldots(x-x_{2015}) = 5\displaystyle \prod_{i=1}^{2015} (x-x_i)$

$P(-1) = 5\displaystyle \prod_{i=1}^{2015}(-1-x_i) = 5 (-1)^{2015} \prod_{i=1}^{2015} (1+x_i)$

$-\dfrac{P(-1)}{5} = \prod_{i=1}^{2015} (1+x_i)$

$\Rightarrow P'(x) = 5(1.(x-x_2).(x-x_3)\ldots(x-x_{2015}) + (x-x_1).1.(x-x_3)\ldots(x-x_{2015}) + \ldots + (x-x_1).(x-x_2)\ldots(x-x_{2014}).1)$

$\displaystyle \sum_{i=1}^{2015} \dfrac{1}{1-x_i} = \displaystyle \dfrac{5[(1-x_2).(1-x_3)\ldots(1-x_{2015}) + (1-x_1).(1-x_3)\ldots(1-x_{2015}) + \ldots + (1-x_1).(1-x_2)\ldots(1-x_{2014})]}{5 \prod_{i=1}^{2015} (1 - x_i)}$

$\Rightarrow \displaystyle \sum_{i=1}^{2015} \dfrac{1}{1-x_i} = \dfrac{P'(1)}{P(1)}$

The net expression is now,

$E = \dfrac{- P(-1)}{5} \dfrac{P'(1)}{P(1)} = \dfrac{4}{5} \dfrac{10078}{8} = \dfrac{10078}{10} = \boxed{1007.8}$

$1+x_i$ 's are the roots of $P(x-1)=5(x-1)^{2015}+x^2-x+1$ $1-x_i$ 's are the roots of $P(1-x)=-5(x-1)^{2015}+x^2-3x+3$ Now $\prod_{i=1}^{2015} (1+x_i)$ is the product of roots of $P(x-1)$ , which equals to $(-1)^{2015} \frac{-5+1}{5}=\frac{4}{5}$ And using Vieta's formulas for $P(1-x)$ we get $\sum_{i=1}^{2015} \frac{1}{1-x_i}=\frac{\sum_{i=1}^{2015} \left(\frac{\prod_{j=1}^{2015} (1-x_j)}{(1-x_i)}\right)}{\prod_{i=1}^{2015} (1-x_i)}=\frac{(-1)^{2014} \frac{-5\times2015-3}{-5}}{(-1)^{2015} \frac{5+3}{-5}}=\frac{10078}{8}$ So the answer is $\frac{4}{5}\frac{10078}{8}=1007.8$

$x_i$ are roots of $P(x),i=1,2,...,2015$

So $(1+x_i)$ are roots of $P(x-1)=5(x-1)^{2015}+(x-1)^2+(x-1)+1$

Clearly the product $\displaystyle \prod_{i=1}^{2015} (1+x_i)=-\frac{\text{constant term}}{\text{coefficient of }x^{2015}} = -\frac{(-5+1-1+1)}{5}=\frac{4}{5}$

Now $P(x)=\displaystyle \prod_{i=1}^{2015} (x-x_i)$ Differentiation yields $P'(x)=P(x)\left(\displaystyle \sum_{i=1}^{2015} \frac{1}{x-x_i} \right )$

Put $x=1$

$\large \left(\displaystyle \sum_{i=1}^{2015} \frac{1}{1-x_i} \right )=\frac{P'(1)}{P(1)}=\frac{2015 \times 5+2+1}{8}=\frac{10078}{8}=\frac{5039}{4}$

Multiplying both we get

Answer= $\large \frac{5039}{4} \times \frac{4}{5}=\boxed{1007.8}$