Player B, who tries to force a real root, needs only to ensure that $ P(t) \leq 0$ for some $ t \in \mathbb{R}$. This is because the leading coefficient $ x^{2n}$ forces the polynomial to tend to infinity when $ x \to \infty$, and in particular it takes on positive values; if takes a negative value anywhere it must also vanish at some point, being continuous. We will show that B indeed has the winning strategy, which is somewhat surprising given that he makes neither the first nor the last move. Since it's difficult to analyze the behavior of the roots as the polynomial is being filled, we'll take the lazy route and only consider B's last move which occurs when there are only two unfilled coefficients - say, the coefficients of $ a^k$ and $ a^l$. So at this point in the game we have $ P(x) = Q(x) + a_k x^k + a_l x^l$ where $ P(x)$ is the polynomial that the players are constructing, $ Q(x)$ is the part of it which is already filled in and is therefore known at this stage, and $ a_k$ and $ a_l$ are the last two remaining coefficients. Player B gets to choose which one he wants to determine and which value to give it, and subsequently player A will have the last word and determine the last coefficient. First, suppose that one of $ k,l$ is even and the other is odd. Say $ k$ is odd and $ l$ is even. Notice that $ P(1) = Q(1) + a_k + a_l$ $ P( - 1) = Q( - 1) - a_k + a_l$ so $ P(1) + P( - 1) = Q(1) + Q( - 1) + 2a_l$ therefore, by choosing an appropriate value for $ a_l$, player B can force $ P(1) + P( - 1)$ to be 0 no matter what player A will then do with $ a_k$. So he does that, and then necessarily either $ P(1) \leq 0$ or $ P( - 1) \leq 0$ which, as we saw, is sufficient for him to be the winner. Next, suppose that both $ k$ and $ l$ are odd. Once again we're trying to find a positive linear combination of values of $ P(x)$ such that one of the coefficients $ a_k$, $ a_l$ has no influence. We can't do this with $ P(1)$ and $ P( - 1)$ so we try $ P(2) = Q(2) + 2^k a_k + 2^l a_l$ $ P( - 1) = Q( - 1) - a_k + a_l$ so now $ P(2) + 2^kP( - 1)$ is independent of $ a_k$, indeed $ P(2) + 2^kP( - 1) = Q(2) + 2^kQ( - 1) + (2^l + 2^k)a_l$ so player B can choose $ a_l$ appropriately and force $ P(2) + 2^kP( - 1)$ to be zero. As before, player A can no longer change that fact, and as before it guarantees that at least one of $ P(2)$ or $ P( - 1)$ is non-positive, which implies a real root. The last case is when both $ k$ and $ l$ are even. Here it's not clear how to find a "good" linear combination as above. However, all's not lost: All player B needs to do is to make sure during the game that at least one of the last two remaining coefficients is odd. This is possible since at the beginning the number of odd coefficients exceeds the number of even coefficients by 1: we have $ n$ odd coefficients and $ n - 1$ even ones. Whatever coefficient $ A$ picks on the first move, B can ensure that this surplus is maintained (if A picks even, B can pick whatever, and if A picks odd, B picks even), and he can obviously continue to do so on subsequent moves. So B ensures that at least one of the last two remaining coefficients is of an odd power of $ x$ and we're done. Note that it doesn't matter which values B actually chooses to place for the coefficients until his very last move.

$B$ has a winning strategy. Since $B$ has the last turn, Let the sum of the coefficients of all the filled spaces be $z$ (including that of $x^{2n}$ and $1$), then $B$ simply fills $-z$ as his last number. And now since sum of the coefficients of the polynomial is $0$, Pretty obviously $P(1)=0$ which is a real root of this equation and thus we are done.