The polynomial $f(x)=x^4+ax^3+bx^2+cx+d$ has real coefficients, and $f(2i)=f(2+i)=0$.  What is $a+b+c+d$?
I notice that the polynomial has real coefficients, which means that complex roots must come in conjugate pairs. Since $f(2i)=0$, I know that $f(-2i)=0$ as well, by the conjugate root theorem. Similarly, since $f(2+i)=0$, I know that $f(2-i)=0$ as well. Therefore, the polynomial has four roots: $2i, -2i, 2+i, 2-i$. I can use these roots to factor the polynomial as follows: $f(x)=(x-2i)(x+2i)(x-2-i)(x-2+i)$. I can simplify this factorization by multiplying the first two factors and the last two factors: $f(x)=(x^2+4)(x^2-4x+5)$. I can expand this product to get the standard form of the polynomial: $f(x)=x^4-4x^3+9x^2-16x+20$. Comparing this with the given form of the polynomial, I can see that $a=-4, b=9, c=-16, d=20$. Therefore, $a+b+c+d=-4+9-16+20=9$.

