Welcome to our community

Be a part of something great, join today!

[SOLVED] Evaluate the constant in polynomial function

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,601
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 $f$.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
384
If [tex]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)[/tex] then obviously, [tex]f= x_1x_2x_3x_4x_5x_6x_7x_8[/tex] so it is a matter of determining possible values for the "x"s. You also know that the coefficient of [tex]x^7[/tex] is [tex]x_1+ x_2+ x_3+ x_4+ x_5+ x_6+ x_7+x_8= -4[/tex], etc.
 
  • Thread starter
  • Admin
  • #2

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,601
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?
 

kaliprasad

Well-known member
Mar 31, 2013
1,283
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

$0 < f<= \frac{1}{2^8}$
 
  • Thread starter
  • Admin
  • #4

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,601
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.
 

kaliprasad

Well-known member
Mar 31, 2013
1,283
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 $\sqrt[8]2$
 

kaliprasad

Well-known member
Mar 31, 2013
1,283
2nd term 7. not 28
 

kaliprasad

Well-known member
Mar 31, 2013
1,283
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 $f = \frac{1}{\sqrt[8]2}$