Answer: B

Solution:

If all three coefficients had one value $u,$ the polynomial would be $u(x^2+x+1),$ whose roots do not equal $u.$ Thus the coefficient and root sets both have two distinct values, so exactly two coefficients coincide. By Vieta's formulas, $r+s=-\dfrac{b}{a}$ and $rs=\dfrac{c}{a}.$

First suppose $a=b=u$ and $c=v.$ The roots are $u,v,$ so Vieta gives $u+v=-1$ and $uv=v/u.$ Hence $v(u^2-1)=0,$ producing $x^2+x-2$ and $-x^2-x.$

If $b=c=v$ and $a=u,$ the same product equation gives $v(u^2-1)=0,$ while the sum equation leaves only $u=1,\ v=-\tfrac12.$ This gives $x^2-\tfrac12x-\tfrac12.$

Finally, if $a=c=u$ and $b=v,$ the product equation gives $uv=1,$ so $v=1/u.$ The sum equation becomes $u^3+u+1=0.$ This strictly increasing cubic has one real root $u,$ producing exactly one more polynomial, $ux^2+\dfrac1u x+u.$ Therefore there are $4$ polynomials.

Thus, B is the correct answer.