Non-constant polynomial factors with leading coefficient 1

Question

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

Answer 1

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?

Answer 2

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 $2^{2015}-1$.