What Is the Product of (1+k) for All Roots of a 15th Degree Polynomial?

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SUMMARY

The product of (1+k) for all roots of the 15th degree polynomial defined by the equation x15-2x14+3x13-...+15x-16=0 is evaluated using Vieta's formulas. The roots k1, k2, ..., k15 yield the result of 16 when calculated as (1+k1)(1+k2)...(1+k15). This conclusion is supported by the contributions of forum members castor28, lfdahl, and kaliprasad, who provided correct solutions to the problem.

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Here is this week's POTW:

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If $k_1,\,k_2,\,\cdots,\,k_{15}$ are the roots of the equation $x^{15}-2x^{14}+3x^{13}-\cdots+15x-16=0$, evaluate $(1+k_1)(1+k_2)\cdots(1+k_{15}).$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution: (Smile)

1. castor28
2. lfdahl
3. kaliprasad

Solution from castor28:
If $f(x)$ is the polynomial, the factors $(1+k_i)$ are the roots of $g(x)=f(x-1)$.

The product of these roots is minus the constant term of $g(x)$:
$$\begin{align*}
-g(0) &= -f(-1)\\
&= -(-1 - 2 + \cdots - 16)\\
&= 136
\end{align*}
$$
 

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