100 Roots

$p(x) = x^{100} - 5x +2\\ p(x) = 0, \text{ for } x = r_1, r_2, \ldots r_{100} \\ p(r_1) = 0 \\ r_1^{100} - 5r_1 + 2 = 0 \\ \implies r_1^{100} = 5r_1 - 2 \\ p(r_n) = 0 \\ r_n^{100}-5r_n+2 = 0 \\ r_n^{100} = 5r_n - 2 \\ \text{Sum of }r_1^{100} + r_2^{100} + \ldots + r_{100}^{100} \\ = \underbrace{5r_1 - 2 + 5r_2 - 2 + \ldots + 5r_{100} - 2}_{\text{100 terms}} \\ = \underbrace{5r_1 + 5r_2 + \ldots + 5r_{100}}_{\text{100 times}} + \underbrace{-2 -2 -2 -\ldots -2}_{\text{100 times}} \\ = 5(r_1 + r_2 + r_3 + \ldots r_{100}) + 100 \times (-2) \\ = 5(0) - 200 ~~ \boxed{\text{Using Vieta's formula sum of zeros of a polynomial = } \dfrac{-b}{a} \text{, here b=0, so sum of zeros is 0}} \\ \boxed{= -200}$

Define that for roots of a polynomial with complex coefficients $r_{1}, r_{2}, r_{3},\cdots,r_{n}$ , $P_{k}=r_{1}^{k}+r_{2}^{k}+r_{3}^{k}+\cdots+r_{n}^{k}$ and $S_{k}$ denotes the $k$ -th symmetric sum of the roots. P_{100} is what we want to find. We use Newton's Sums for the roots: $P_{100}=S_{1}P_{99}-S_{2}P_{98}+S_{3}P_{97}-S_{4}P_{96}+\cdots-S_{98}P_{2}+S_{99}P_{1}-100S_{100}$ . From Viete's Formula we know $P_{1}=S_{1}=S_{2}=S_{3}=\cdots=S_{97}=S_{98}=0$ and $S_{99}=5, S_{100}=2$ Substituting the values, $P_{100}=-100S_{100}=-200$