Very Necessarily Large Powered Polynomial

The non-zero roots of $P(x)$ is given by: $4x^3-4x^2+x+7 = 0$ . Let the roots be $a$ , $b$ and $c$ . By Vieta formulas, we have:

$a+b+c = 1\quad \quad ab+bc+ca = \frac {1}{4}\quad \quad abc = -\frac {7}{4}$

Using Newton Sums method, we have:

$M = a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 1^2-\frac {2}{4} = \frac {1}{2}$

$N = a^3+b^3+c^3$

$\quad = (a+b+c)(a^2+b^2+c^2) - (a+b+c)(ab+bc+ca) + 3abc$

$\quad = 1(\frac{1}{2})- \frac {1}{4}(1) + 3(-\frac {7}{4}) = \dfrac {2-1-21}{4} = -5$

$\Rightarrow M\times N = \frac {1}{2} \times (-5) = \boxed{-2.5}$

First of all, we must try to define the sum of the squares of the roots in a polynomial and remember of Vieta's formula:

$\displaystyle \frac{1}{1!}\sum_{i=1}^{n} {x_i}=\frac{-b}{a}$ ,

$\displaystyle \frac{1}{2!}\sum_{i=1,j=1,{i}\neq{j}}^{n} {x_i}.{x_j}=\frac{c}{a}$ and

$\displaystyle \frac{1}{3!}\sum_{i=1,j=1,k=1,{i}\neq{j}\neq{k}}^{n} {x_i}.{x_j}.{x_k}=\frac{-d}{a}$

where $n$ stands for the degree of the polynomial, $a$ stands for the first coefficient, $b$ stands for the second coefficient, $c$ stands for the third coefficient and $d$ stands for the fourth coefficient. The square of a sum can be expressed as follows:

$\displaystyle(\sum_{i=1}^{n} {x_i})^{2}=\displaystyle \sum_{i=1}^{n} {x_i}^{2}+\displaystyle {\sum_{i=1,j=1,{i}\neq{j}}^{n} {x_i}.{x_j}}$

With some substitutions, we have:

$(\frac{-b}{a})^{2}=M+\frac{2c}{a}\rightarrow M=(\frac{b}{a})^{2}-\frac{2c}{a}$

By the same way, we try to set the sum of cubes, $N$ :

$\displaystyle(\sum_{i=1}^{n} {x_i})^{3}=\displaystyle \sum_{i=1}^{n} {x_i}^{3}+\displaystyle {\sum_{i=1,j=1,{i}\neq{j}}^{n} {x_i}^{2}.{x_j}}+\sum_{i=1,j=1,k=1,{i}\neq{j}\neq{k}}^{n} {x_i}.{x_j}.{x_k}$

$(\frac{-b}{a})^{3}=N+\displaystyle {\sum_{i=1,j=1,{i}\neq{j}}^{n} {x_i}^{2}.{x_j}}-\frac{6d}{a}$

Now, we need to define the sum left. As $j$ varies from $1$ to $n$ , every root is "taken" but one, that is the one defined by $i$ . Hence, the summation assumes the form below:

$\displaystyle \frac{2!}{3!}{\sum_{i=1,j=1,{i}\neq{j}}^{n} {x_i}^{2}.{x_j}}={x_1}^{2}.({x_2}+{x_3}+...+{x_n})+{x_2}^{2}.({x_1}+{x_3}+...+{x_n})+...+{x_n}^{2}.({x_1}+{x_2}+{x_3}+...+{x_{n-1}})=$

$={x_1}^{2}.(\frac{-b}{a}-{x_1})+{x_2}^{2}.(\frac{-b}{a}-{x_2})+...+{x_n}^{2}.(\frac{-b}{a}-{x_n})$

$=\frac{-b}{a}.({x_1}^{2}+{x_2}^{2}+...+{x_n}^{2})-({x_1}^{3}+{x_2}^{3}+...+{x_n}^{3})$

$=\frac{-b.M}{a}-N$

And back to the sum of cubes, we'll have:

$(\frac{-b}{a})^{3}=N+3.[\frac{-b.M}{a}-N]-\frac{6d}{a} \rightarrow N=(\frac{-1}{2}).[(\frac{-b}{a})^{3}+\frac{3bM}{a}+\frac{6d}{a}]$

From now on, we've just got to compute $M$ and $N$ :

$M=(\frac{-4}{4})^{2}-\frac{2.1}{4}=\frac{1}{2}$

$N=(\frac{-1}{2}).[(\frac{-(-4)}{4})^{3}+\frac{3.(-4).\frac{1}{2}}{4}+\frac{6.7}{4}]=-5$

Thus, $M \times {N} = \boxed{-2.5}$

Let $P_n$ be the sum of the power of n of its roots. By Newton's Sum, we provide:

$4P_1-4=0$

$4P_2-4P_1+2=0$

$4P_3-4P_2+P_1+21=0$

Solving each variable, we provide:

$P_1=x_{1}+x_{2}+...+x_{79}=1$

$P_2=x_{1}^2+x_{2}^2+...+x_{79}^2=0.5$

$P_3=x_{1}^3+x_{2}^3+...+x_{79}^3=-5$

The solution of $M$ is $P_2=0.5$ while the solution of $N$ is $P_3=-5$ . Finally, $M \times N = 0.5 \times -5 = -2.5$