Here is my solution. I am not sure if it is right or not, so please inform me of this. The answer is that the opponent cannot spoil his efforts. Here is the first pupil's strategy: First he assigns some positive real number to the constant term. Now his opponent chooses either the $ x^2$ or $ x$ term to attach a coefficient to. This leaves the first pupil with the task of choosing a real number $ a$ such that $ p(x)=ax^2$ or $ p(x)=ax$, depending on what the opponent did previously, has only one real solution, where $ p(x)$ is a cubic with all the chosen coefficients. Now we know for $ p(x)$ the relative minimum occurs to the right of the y-axis, since $ p(0)>0$ and $ p(x)\to \infty$ as $ x\to \infty$ and vice versa. If the first pupil must choose a real $ a$ for the equation $ p(x)=ax$ to have one unique real solution, he does this by taking $ a$ to be negative so that we know the line intersects the cubic at some point $ (x_1,y_1)$ with $ x_1<0$. The pupil also must make $ |a|$ large enough so that the line $ y=ax$ does not intersect the cubic at on the interval between $ x_1$ and the relative minimum. This way, the line only intersects $ p(x)$ once. The pupil works similarly in the case of $ p(x)=ax^2$ having a unique real solution. First, the pupil chooses $ a$ to be negative so that we know that the parabola and the cubic intersect at some point $ (x_1,y_1)$ for $ x_1<0$ (since the cubic decreases faster than the parabola for $ x<0$). The pupil also chooses $ |a|$ to be large enough so that the parabola does not intersect the cubic on the interval between $ x_1$ and the relative minimum.

also works if the first pupil assigns at the constant term 0????? the second pupl should to assign 0 to the second dots. Here is the first pupil's strategy: the first pupil assign 0 at the first dots, if the second pupil assigns at the constant term 'a' then the first pupil chooses 1, therefore P'(x)=3x*x+1>0 ( P is increasant) since degree(P)=3:odd then P(- \infinite )=- \infinite and P( \infinite)= \infinite hence P has exactly one real root; if the secondpupil assign 0 at the seconddots, the first pupil chooses 1, so -1 is the one real root of P if the second pupil chooses 'a' >0 therefore P'(x)=3x*x+a>0(P is increasant) since degree(P)=3:odd then $ P(-infinite)=-infinite$ and $ P(infinite)=infinite$ hence P has exactly one real root; if the second pupil chooses '-a'; a>0 then the first pupil choose $ b>(2a/9) *\sqrt{3a}$ so the local minimum is positive, and p has one real root.