Evaluate the constant in polynomial function

In summary, the polynomial with given factors has a coefficient of $-4$ for $x^7$ and a coefficient of $7$ for $x^6$. The value of $f$ can be determined by multiplying all the factors together, and the value of $-4$ can be found by adding all the factors. The other coefficients of the polynomial are not relevant in determining the value of $f$.
  • #1
anemone
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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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  • #2
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.
 
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  • #2
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?
 
  • #3
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

$0 < f<= \frac{1}{2^8}$
 
  • #4
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.
 
  • #5
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 $\sqrt[8]2$
 
  • #5
2nd term 7. not 28
 
  • #6
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 $f = \frac{1}{\sqrt[8]2}$
 

1. What is a constant in a polynomial function?

A constant in a polynomial function is a number that does not have a variable attached to it. It is typically represented by the letter "c" and is added or subtracted to the variable terms in the function.

2. How do you evaluate the constant in a polynomial function?

To evaluate the constant in a polynomial function, you simply plug in the given value for the variable and solve the resulting equation. This will give you the specific value of the constant in that particular function.

3. Can the constant in a polynomial function be negative?

Yes, the constant in a polynomial function can be negative. This means that when the variable is equal to 0, the overall value of the function will be negative.

4. What is the purpose of the constant in a polynomial function?

The constant in a polynomial function helps to shift the graph of the function up or down on the y-axis. It also helps to determine the y-intercept of the function.

5. Is the constant always necessary in a polynomial function?

No, the constant is not always necessary in a polynomial function. If the constant is equal to 0, it does not affect the shape or behavior of the function. However, it is still important to include the constant in the function to accurately represent the equation.

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