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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.

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  • $\begingroup$ resulting formula has atleast one zero or exactly one zero? $\endgroup$ – Sagar Chand Oct 21 '17 at 10:01
  • $\begingroup$ Exactly one real zero. $\endgroup$ – Larry Powell Oct 21 '17 at 10:01
  • $\begingroup$ Can you explain the rules of the game ? $\endgroup$ – Yves Daoust Oct 21 '17 at 10:02
  • $\begingroup$ Given the polynomial in the original post, we take turns given a value to the coefficients. Such as Player 1 chooses $a=1$ Player 2 chooses $d=0$ Player 1 chooses $c=-1$ No matter what player 2 chooses the polynomial will have two zeroes. Therefore player 1 wins. However if player two manages to have just one zero in the resulting polynomial he will win. $\endgroup$ – Larry Powell Oct 21 '17 at 10:08
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    $\begingroup$ Is a double root counted as one or two ? $\endgroup$ – Yves Daoust Oct 21 '17 at 10:25
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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.

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    $\begingroup$ one weak spot: if player 1 begins with $b = 1$, we choose $d = 1$. Then if player 1 chooses $0$ for either of the others, according to this strategy we should choose to make $a = c = 0$, but this gives a polynomial with no real roots. Easily fixable, though, I think. $\endgroup$ – Nick Pavlov Oct 21 '17 at 12:20
  • $\begingroup$ @NickPavlov you are right, I will try to fix this. $\endgroup$ – Reinier Oct 21 '17 at 12:54
  • $\begingroup$ I have now fixed the weak spot, thanks for spotting it! $\endgroup$ – Reinier Oct 21 '17 at 13:09

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