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

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To evaluate the product (1+k1)(1+k2)...(1+k15) for the roots of the polynomial x^15 - 2x^14 + 3x^13 - ... + 15x - 16 = 0, one can use Vieta's formulas. The expression can be rewritten as P(1), where P(x) is the polynomial. By substituting x = 1 into the polynomial, the value of P(1) is calculated. The final result of the product is found to be 16. This demonstrates a method for evaluating products of shifted roots in polynomial equations.
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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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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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