Given $a_n∈\mathbb{Z}$ with $a_{10}=11$ and $a_9=-143$, determine the number of polynomials of the form $$P(x)=\sum_{i=0}^{10} a_nx^n$$ such that the zeroes of $P(x)$ are all positive integers
I've encountered this question, and I don't really know what it means by polynomials of the given form. Isn't $P(x)$ already distinct given the sequence? I'm probably grossly misunderstanding something here.
Let all roots of $P(x)$ be $b_1,b_2,\cdots,b_{10}$ which are positive integers with $1\le b_1\le b_2\le \cdots\le b_{10}$. Then $$ P(x)=\sum_{i=0}^{10}a_ix^i=11\prod_{i=1}^{10}(x-b_i). $$ Comparing the constant terms, one has $$ a_0=11\prod_i^{10}b_i $$ which implies $a_0=11k$ ($k$ is positive integer). Comparing the coefficients of $x^9$, one has $$ \sum_{i=1}^{10}b_i=\frac{143}{11}=13. $$ So $$ 10b_1\le 13 $$ from which one has $$ b_1=1$$ Similarly $b_2=\cdots=b_7=1$. So $$ b_8+b_9+b_{10}=6. $$ It is easy to see $b_8=b_9=b_{10}=2$. Therefore there is only one polynomial satisfying the conditions.