Which player wins the game of cubic coefficients forcing one zero. The players are given 
$ ax^3 + bx^2+ cx+ d.$
Players alternate taking turns putting in a real coefficient, including zero.
Player 2 will win if the resulting formula has exactly one real zero, else player 1 will win.
I know at least player 1 cannot let player 2 have the constant last, his first move also cannot be $a=0$ and that if player 1 takes $a \ne 0$ and player 2 sets $d=0$ player 1 can win by $sign(c)=-sign(a)$ . However I feel like I'm looking at too many options and not zeroing at the right solution.
 A: Player 2 has a winning tactic. This player can namely force a polynomial of the form $ax^3+bx^2+ax+b = (x^2+1)(ax+b)$ with $a \neq 0$, $bx^2+2bx+b = b(x+1)^2$ with $b \neq 0$, $2bx^3+bx^2+b = b(2x^2-x+1)(x+1)$ with $b \neq 0$, or $e\cdot x^i$ with $i > 0$. All these polynomials have exactly one real zero. This can be done with the following tactic:
Suppose that player 1 starts with setting one coefficient to $0$. Now we apply the following tactic:
First move:


*

*If player 1 set $d=0$, player 2 can take $a=0$.

*If player set some other coefficient to $0$, player 2 takes $d=0$.
Second move:


*

*If player 1 set some other coefficient to $0$, player 2 sets the remaining coefficient to $1$, resulting in a polynomial $x^i$ for some $i >0$ (because we forced $d = 0$ in the first move).

*If player 1 does something else, then player 2 sets the remaining coefficient to $0$, which results in some polynomial $e\cdot x^i$ for some $i > 0$.
In the case that the first player did not start with a $0$, player 2 can do the following:
First move:


*

*If player 1 picked $a$ or $c$, player 2 can take the other number of these two in such a way that $a = c$.

*If player 1 picked $b$ or $d$, player 2 can also take the other number of these two to get $b = d$
Second move:


*

*If in the first move we enforced that $a = c$, we can now easily take $b = d$. This results in a polynomial of the form $ax^3+bx^2+ax+b$ with $a \neq 0$.

*Similarly, if we had $b = d$ after the first move, and player one did not take $a = 0$ or $c = 0$, we can now pick the remaining coefficient such that $a = c$, and again the result is of the form $ax^3+bx^2+ax+b$ with $a \neq 0$.

*If the player 1 made the move $a = 0$, we set $c = 2b$, resulting in the polynomial $bx^2+2bx+b$ with $b \neq 0$.

*Finally, if player 1 put $c = 0$, then we set $a = 2b$, giving the polynomial $2bx^3+bx^2+b$ with $b \neq 0$.
So we see that this tactic always results in a polynomial with exactly one zero.
