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How many (nonconstant) polynomial factors with leading coefficient 1, with the other coefficients possibly complex, does $x^{2015} + 18$ have?

I don't know much about this problem, all I know is that the complex roots must come in pairs? I'm not sure where to read up to be able to solve this problem

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  • $\begingroup$ I think the only real issue to resolve here is whether there could be repeated factors, since all real polynomials split when factored over the complex numbers. To find any repeated factors, compute the GCD with the derivative of the polynomial. $\endgroup$ – hardmath Aug 14 '18 at 15:53
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Let $p(x)$ be your polynomial. It has $2\,015$ roots: the numbers $r_j=\sqrt[2\,015]{-18}\exp\left(\frac{2\pi ij}{2\,015}\right)$ with $j\in\{0,1,\ldots,2\,014\}$. So, it's monic factors are the polynomials of the form$$\prod_{j\in\Delta}(x-r_j),$$where $\Delta$ is a non-empty subset of $\{0,1,\ldots,2\,014\}$. How many non-empty subsets has this set?

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  • $\begingroup$ Im sorry, i dont get the 1st equation wihh rj, can you link me a source to where i can read it? This is a new concept 2 me $\endgroup$ – SuperMage1 Aug 14 '18 at 15:54
  • $\begingroup$ Im sorry, i dont get the 1st equation wihh rj, can you link me a source to where i can read it? This is a new concept 2 me $\endgroup$ – SuperMage1 Aug 14 '18 at 15:54
  • $\begingroup$ What I wrote was he $2\,015^{\text{th}}$ root of $-18$ times the exponencial of $frac{2\pi\times i\times j}{2\,015}$. Is that your problem? You were unable to read the expression? $\endgroup$ – José Carlos Santos Aug 14 '18 at 15:59
  • $\begingroup$ I was unable make sense of that equation and how you got it. $\endgroup$ – SuperMage1 Aug 14 '18 at 16:07
  • $\begingroup$ I want to understand it but i do not know the nmae of the concept as 8m very new to this $\endgroup$ – SuperMage1 Aug 14 '18 at 16:08
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By the fundamental theorem of algebra the polynomial is a product of linear factors with complex coefficients. That is to say, there exist $\alpha_1,\ldots\alpha_k\in\Bbb{C}$ and positive integers $m_1,\ldots,m_k$ such that $$x^{2015}+18=\prod_{j=1}^k(x-\alpha_i)^{m_i}.$$ Here the $\alpha_i$ are the roots of $x^{2015}+8$ and the $m_i$ are their multiplicities. Now the polynomial factors of $x^{2015}+18$ are precisely the different products of linear factors, each with multiplicity at most $m_i$. To count the number of different products, we first count the number of distinct linear factors $k$.

Comparing degrees shows that $\sum_{j=1}^km_i=2015$, where $m_i\geq1$ for all $i$. In particular we already see that $k\leq2015$. Now suppose that $k<2015$. Then there must be a root $\alpha_i$ with multiplicity $m_i>1$ for the equality $\sum_{j=1}^km_i=2015$ to hold. But then the derivative (computed using the product rule) $$\frac{d}{d x}\prod_{j=1}^k(x-\alpha_i)^{m_i}=\sum_{j=1}^k\left(m_i(x-\alpha_j)^{m_i-1}\prod_{\substack{\ell=1\\\ell\neq j}}^k(x-\alpha_\ell)^{m_\ell}\right),$$ has a linear polynomial factor $x-\alpha_i$, so $\alpha_i$ is also a zero of the derivative. But of course the derivative equals $$\frac{d}{d x}(x^{2015}+18)=2015x^{2014},$$ with $0$ as its only zero, which clearly is not a zero of $x^{2015}+18$, a contradiction. Hence $k=2015$, there are $2015$ distinct linear factors of $x^{2015}+18$.

Now the number of different products is easy to count; every such product is determined by whether it contains the factor $x-\alpha_i$ or not, for each $i$. This yields precisely $2^{2015}$ polynomial factors with leading coefficient $1$, of which only one (corresponding to the empty product) is constant. So the answer is $2^{2015}-1$.

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  • $\begingroup$ @hardmath Oh my, you're absolutely right! I'll edit my answer accordingly. $\endgroup$ – Servaes Aug 15 '18 at 0:44
  • $\begingroup$ Thanks, I understand it, but the answer is not 2^2015, but 2^2015-1, Where could that come from? $\endgroup$ – SuperMage1 Aug 15 '18 at 10:56
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    $\begingroup$ does it have to do with the factor being non constant? $\endgroup$ – SuperMage1 Aug 15 '18 at 10:57
  • $\begingroup$ Yes, it is precisely that. I forgot about the empty product, which is constant. I have edited accordingly. $\endgroup$ – Servaes Aug 15 '18 at 14:09

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