MHB Evaluate the constant in polynomial function

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The polynomial P(x) is expressed as a product of eight linear factors, with the condition that all roots are positive real numbers. The constant term f is equal to the product of the roots, x_1x_2x_3x_4x_5x_6x_7x_8, and the sum of the roots must equal -4, as indicated by the coefficient of x^7. Participants in the discussion are tasked with determining the possible values of f, with one contributor asserting that there is only one valid solution. Clarifications regarding the coefficients and terms of the polynomial are also addressed, emphasizing the importance of accurate calculations. The conversation revolves around finding the precise value of f based on the established relationships among the polynomial's coefficients and roots.
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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 $f$.
 
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If P(x)= (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)(x-x_8) then obviously, f= x_1x_2x_3x_4x_5x_6x_7x_8 so it is a matter of determining possible values for the "x"s. You also know that the coefficient of x^7 is x_1+ x_2+ x_3+ x_4+ x_5+ x_6+ x_7+x_8= -4, etc.
 
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Actually I do know a lot! But, since this is a challenge problem, what I expect (like how I told HallsofIvy before) is the full solution along with the logical explanation. Okay?
 
Using Vieta's formula we have

$\sum_{1=1}^8 x_i = 4\dots(1)$
$\sum_{1=1}^7 \sum_{j=i+1}^8 x_ix_j = 7\cdots(2)$
$\prod_{1=1}^8 x_i = f\cdots(3)$From (1) and (3) using AM GM inequality as all are positive taking AM ad GM of the elements of parameters of equation (1)
$AM = \frac{\sum_{1=1}^8 x_i}{8} = \frac{4}{8} = \frac{1}{2}\dots(4)$
$GM= \sqrt[8]{\prod_{1=1}^8x_i}=\sqrt[8] f\cdots(5)$
As AM is larger than equal to GM we get from (4) and (5)
$f<= \frac{1}{2^8}\cdots(6)$

From (2) and (3) using AM GM inequality taking AM ad GM of the elements of parameters of equation (2)
$AM = \frac{\sum_{1=1}^7 \sum_{j=i+1}^8 x_ix_j }{28}= \frac{7}{28} = \frac{1}{4}\cdots(7)$ as there are 28 terms
$GM = \sqrt[28]{\prod_{1=1}^7 \prod_{j=i+1}^8 x_ix_j }= \sqrt[28]{(\prod_{1=1}^8)^7}= \sqrt[28]{f^7}= \sqrt[4]{f} \cdots(8)$
As AM is larger than equal to GM we get from (7) and (8)
$f<= \frac{1}{4^4})$
or $f<= \frac{1}{2^8}\cdots(9)$

From (6) and (9) we have $f<= \frac{1}{2^8}$

Actually f becomes $\frac{1}{2^8}$when all the $x_i$ are $\frac{1}{2}$

As any $x_i$ can be as low as possible but not zero we we get the condition

$0 < f<= \frac{1}{2^8}$
 
Hi kaliprasad, thanks for participating but your answer is not quite right because there is only one value for $f$. I will give others a chance to take a stab at it, before I post the solution.
 
Hello anemone,-

Thanks. I realize the mistake. I realized that though the range is calculated for for both cases this does not mean for same $x_i$ which meets the sum to be 4 the 2nd term shall be 28. This I can not prove or disprove it. But if tou say that there is only one f then it has to be $\sqrt[8]2$
 
2nd term 7. not 28
 
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 $f = \frac{1}{\sqrt[8]2}$
 
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