AMC12 2020 B

Answer (C): Because any such polynomial has constant term $2020\neq 0$, the polynomial must have five nonzero roots whose product is $-2020$. Because the polynomial has real coefficients, all nonreal roots appear in complex-conjugate pairs. Because the polynomial has exactly five roots, there must be a real root $s\neq 0$ such that each root of the polynomial is one of $$ s,\qquad \alpha=\frac{-1+i\sqrt3}{2}\cdot s,\qquad \text{or}\qquad \overline{\alpha}=\frac{-1-i\sqrt3}{2}\cdot s, $$ for otherwise there would be at least six distinct roots. Because the product of $\alpha$ and $\overline{\alpha}$ is $s^2$, it follows that $s=-\sqrt[5]{2020}$. Either $\alpha$ and $\overline{\alpha}$ are both roots of multiplicity $2$, or the complex roots each have multiplicity $1$ and $s$ has multiplicity $3$. These are the only possibilities, so there are $2$ such polynomials. Explicitly, they are, respectively, $$ x^5-\sqrt[5]{2020}\,x^4+\sqrt[5]{2020^2}\,x^3+\sqrt[5]{2020^3}\,x^2-\sqrt[5]{2020^4}\,x+2020 $$ and $$ x^5+2\sqrt[5]{2020}\,x^4+\sqrt[5]{2020^2}\,x^3+\sqrt[5]{2020^3}\,x^2+2\sqrt[5]{2020^4}\,x+2020. $$