Probably one of my favourite problems from Sozopol (if not my favourite). Sadly we didn't get the opportunity to present this cool solution (by @GeoGuesrr and me). Here I will show a generalized solution for degree $n$ (where $n$ is even). Denote the polynomial to be $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_{0}$, the polynomial after the $n$-th move to be $P_n(x)$ and the players to be $A$ and $B$. a) Answer: Player B wins. Proof: Case 1: At the last move player B is left to choose $a_m$, where $m$ is odd. Then we want to find $a_m$ s.t. $f(x)=\frac{P_{n-1}(x)}{x^m}=-a_m$. Taking $x\rightarrow -\infty$ we know that $f(x)\rightarrow -\infty$, meaning there $\exists x_1$ s.t. $f(x_1)<0$, so choosing $a_m=-f(x_1)$ we get our desired root. Case 2: Player $B$ is left to choose $a_0$ at the end. We want to find $a_0$ s.t. $P_{n-1}(x)=-a_0$, which is just asking can we find a negative value of $P_{n-1}$. But this is obviously true, because $0$ is a single root of this polynomial. Thus the strategy for Player $B$ is just to delay taking any of the above mentioned coefficients and as they are $\frac{n}{2}+1$, it is possible to insure that he's left with choosing such a value. b) Answer: Player B wins. Proof: We consider $x<0$ as they are no non-negative roots. If Player $B$ is left to choose $a_m$, where $m$ is even. He then wants to insure that for no $x$, $f(x)=\frac{P_{n-1}(x)}{x^m}=-a_m$, which is $\Leftrightarrow$ to proving that $P$ is not surjective. If FTSOC we assume that the last is impossible, then this means that $\sup(f)=-\infty$ in $(-\infty, 0)$. As $f(x)\rightarrow +\infty, x\rightarrow -\infty$ and $f(x)\rightarrow +\infty ,x\rightarrow 0^-$ and $f$ is continuous in our interval it is possible to show that $\sup(f)\neq-\infty$ (Doing this rigorously was quite difficult) from which it follows that we can find a desired value for $a_m$. This means that if $B$ only chooses odd degrees, $A$ must only choose even ones. But now we will show that after an odd number of moves $k$, it is possible for $B$ to choose an odd coefficient $m$ such that if $P_k(x)$ has no roots, then $P_{k+1}(x)$ also has no roots. We want to show that $f(x)=\frac{P_k(x)}{x^m}$ is not surjective for $(0,-\infty)$. We know that $f$ has no roots so by IVT we have either have $\inf(f)<0\Rightarrow f$ is not surjective on our interval or $\inf(f)=0$. Using the fact that $f$ is continuous in $(-\infty,0)$ one can again show that $\inf(f)=0 \Rightarrow f$ has real roots (I'll again omit the details). Thus one can conclude that as $P_0$ has no real root (in our interval), then if $B$ always chooses odd degrees he can insure that no polynomial $P_i$ has a real root (When $A$ adds an even degree he obviously can't make a new real root), from which we are done.