A polynomial $f(x) = x^4 + * x^3 + * x^2 + * x+1$ has three undetermined coefficients denoted by stars. The players A and B move alternately, replacing a star by a real number until all stars are replaced, where A goes first. A wins if all zeros of the polynomial are complex and B wins otherwise. Show that B can always win.
A possible generalization: can one still guarantee that B wins if $f(x) = x^n + *x^{n-1}+\cdots + * x^2 + *x+1$ for arbitrary $n\ge 4$?
I know that the complex zeroes $r_i, 1\leq i\leq n$ of a polynomial $f(x) = a_n x^n+\cdots + a_1x+a_0$ satisfy Vieta's equations; namely $s_j(r_1,\cdots, r_n) = (-1)^{j} \frac{a_{n-j}}{a_n},$ where $s_j$ denotes the jth symmetric polynomial in n variables.
We need to show that for any $a,c \in \mathbb{R}, \exists b\in \mathbb{R}$ so that $ x^4 + ax^3 + bx^2 + cx + 1$ has at least one zero. But how can one prove this? Also, is there a similar proof for the generalization or are there counterexamples?