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Given $a_n$ is an integer with $a_{10} = 11, a_9 = -143$, determine the number of polynomial in the form of

$$P(x) =\sum_{i=0}^{10} a_nx^n$$

such that the zeros of $P(x)$ are all positive integers.

I have never encountered a problem like this, all I know is that $a_{10}= 11$ therefore there is a factor $11x- n$ and $n$ is divisible by $11$, I don't know how to use the $a_9$.

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  • $\begingroup$ By [[Vieta's formulas]] the sum of the roots is equal to $142/11=13$. $\endgroup$ – user583012 Aug 10 '18 at 12:37
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Hint: Let $x_1,...,x_{10}$ be all roots of $P(x)$. Then by Vieta formulas we have

$$ x_1+...+x_{10} = -{a_{9}\over a_{10}} = 13$$

Since $x_1,...,x_{10}$ are all positive integers you can do the stars and bars method now...

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  • $\begingroup$ stars and bars? we cant that because the bins are identical i think, since were getting the disticnt polynomials? so our answer should be 3 I think? $\endgroup$ – SuperMage1 Aug 12 '18 at 9:44
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by Vieta formula we get $$\sum_{i=1}^{10} x_i= -\frac{a_9}{a_{10}} =-\frac{-143}{11}=13$$

For any pair of positive integers n and k, the number of distinct k-tuples of positive integers whose sum is n is given by the binomial coefficient $${n - 1\choose k-1}={12\choose 9} $$.

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