AMC12 2019 B

Answer (B): Because the coefficients are real, so must be the roots. Let the two real roots be $r$ and $s$. Note that $r\neq s$ because $p(x)=rx^2+rx+r$ has no real roots. The given condition implies that, without loss of generality, $p(x)=r(x-r)(x-s)$, with $r\neq 0$. There are three cases to consider. Case 1: $r(x-r)(x-s)=rx^2+rx+s$. Then equating coefficients yields $r+s=-1$ and $r^2s=s$. The solutions to this system of equations are $(r,s)=(1,-2)$, in which case $p(x)=x^2+x-2$, and $(r,s)=(-1,0)$, in which case $p(x)=-x^2-x$. Case 2: $r(x-r)(x-s)=rx^2+sx+r$. Then $r+s=-\frac{s}{r}$ and $rs=1$. Multiplying the first equation by $r^2$ and substituting for $rs$ yields $r^3+r+1=0$. Descartes’ Rule of Signs shows that this equation has no positive real solution and one negative real soution (which is approximately $-0.682$, so $p(x)\approx -0.682x^2-1.466x-0.682$). Case 3: $r(x-r)(x-s)=rx^2+sx+s$. Then $r+s=-\frac{s}{r}$ and $rs=\frac{s}{r}$. It follows from $r\neq 0$ that $s\neq 0$. Then from the second equation, $r^2=1$. If $r=-1$, then the first equation becomes $-1+s=s$, which has no solution. If $r=1$, then the first equation becomes $1+s=-s$, so $s=-\frac{1}{2}$, in which case $p(x)=x^2-\frac{1}{2}x-\frac{1}{2}$. In summary, Case 1 yielded $2$ polynomials, Case 2 yielded $1$ more, and Case 3 yielded $1$ more, for a total of $2+1+1=4$ polynomials.