2021/095) Let $a,\,b,\,c,\,d,\,e,\,f$ be real numbers such that the polynomial

 $P(x)=x^8-4x^7+7x^6+ax^5+bx^4+cx^3+dx^2+ex+f$ factorizes into eight linear factors $x-x_i$ with $x_i>0$ for $i=1,\,2,\,\cdots,\,8$.

Determine all possible values of $f$.

Solution 

Using Vieta's formula  we have

$\sum_{i=1}^8 x_i = 4\dots(1)$

$\sum_{i=1}^7 \sum_{j=i+1}^8   x_ix_j = 7\cdots(2)$

$\prod_{i=1}^8 x_i = f\cdots(3)$

We have

$\sum_{i=1}^8 x_i^2= (\sum_{i=1}^8 x_i)^2 - 2 \sum_{i=1}^7 \sum_{j=i+1}^8   x_ix_j$

$= 4^2 - 2 * 7 = 2$

or $\sum_{i=1}^8 x_i^2 = 2$

Subtracting (1) from above

$\sum_{i=1}^8 (x_i^2 - x_i)  = -2$

adding $\frac{1}{4}$ to each term on LHS that is 2 and adding 2 on RHS we get

$\sum_{i=1}^8 (x_i^2 - x_i + \frac{1}{4})  = 0$

or $\sum_{i=1}^8 (x_i - \frac{1}{2})^2  = 0$

so $x_i = \frac{1}{2}$ for each i.

this satisfies the criteria that $x_i$ is positive

putting this in (3) we get $f = \frac{1}{\sqrt[8]2}$