Extracting Polynomial Parameters

$f(x)=2(x-x_1)(x-x_2)(x-x_3)$ then $g(x)=2(x^2+5x+1-x_1)(x^2+5x+1-x_2)(x^2+5x+1-x_3)$ we want the sum of the squares of roots of each factor, i.e $5^2-2(1-x_1)+5^2-2(1-x_2)+5^2-2(1-x_3)$ $=69+2(x_1+x_2+x_3)$ by vietas formula, the sum is $\frac{-5}{2}$ . so $=69+2(\frac{-5}{2})=69-5=\boxed{64}$

My approach to calculate $\displaystyle \sum_{i} s_{i}^2$ was to use Newton's sums and for doing that we have to calculate first three coefficients of $g(x)$ , i.e., coefficients of $x^6,x^5$ and $x^4$ .

Coefficients of $x^6,x^5$ and $x^4$ in $2(x^2+5x+1)^3$ are $2,30$ and $156$ .

Coefficients of $x^6,x^5$ and $x^4$ in $(x^2+5x+1)^2$ are $0,0$ and $5$ .

Therefore,

$g(x)=2x^6+30x^5+161x^4+p(x)$

Where $p(x)$ is the remaining part of $g(x)$ .

Let $P_{n}=\displaystyle \sum_{i}s_{i}^n$ , so we need to find $P_{2}$ .

By Newton's sums,

$(2)P_{1}+30=0\Rightarrow P_{1}=-15 \\ (2)P_{2}+(30)P_{1}+322=0\Rightarrow P_{2}=\boxed{64}$ .

$f(x)$ can be factored as follows:

$f(x) = 2 (x - r_1)(x - r_2)(x - r_3)$

hence,

$g(x) = 2 (x^2 + 5 x + 1 - r_1)(x^2 + 5 x + 1 - r_2)(x^2 + 5 x + 1 - r_3)$

Hence a solution $s_i$ to g(x) = 0 , satisfies

$s_i^2 + 5 s_i + 1 - r_j = 0$ .. .. .. .. .. Equation (1)

for $i = 1, 2, ...,6$ where $s_1$ and $s_2$ are associated with $r_1$ , etc.

Now, since this is a quadratic equation in $s_i$ , the sum of the two solutions is $-b$ , i.e.

$s_1 + s_2 = s_3 + s_4 = s_5 + s_6 = -5$

Adding the entries of Equation (1) over i=1,2,...6, we obtain

$\sum_i {s_i}^2 + 5 \sum_i {s_i} + \sum_i 1 - \sum_i r_{j(i)} = 0$

where $j(1) = j(2) = 1, j(3) = j(4) = 2 , j(5) = j(6) = 3$ .

Now

$\sum_i {s_i} = 3 (-5) = - 15$

$\sum_i 1 = 6$

$\sum_i r_{j(i)} = 2 (r_1 + r_2 + r_3) = 2 (-5/2) = -5$

Therefore,

$\sum_i s_i^2 = -5(-15) - 6 + (-5) = 75 - 11 = \boxed{64}$