The polynomial $f(x) = x^{4} + ax^{3} + bx^{2} + cx + d$ has real coefficients, and $f(2i) = f(2 + i) = 0.$ What is $a + b + c + d?$
$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 4 \qquad \mathrm{(D)}\ 9 \qquad \mathrm{(E)}\ 16$
A fourth degree polynomial has four [roots](https://artofproblemsolving.com/wiki/index.php/Root). Since the coefficients are real(meaning that complex roots come in conjugate pairs), the remaining two roots must be the [complex conjugates](https://artofproblemsolving.com/wiki/index.php/Complex_conjugate) of the two given roots. By the factor theorem, our roots are $2-i,-2i$. Now we work backwards for the polynomial:

$(x-(2+i))(x-(2-i))(x-2i)(x+2i) = 0$
$(x^2 - 4x + 5)(x^2 + 4) = 0$

$x^4 - 4x^3 + 9x^2 - 16x + 20 = 0$
Thus our answer is $- 4 + 9 - 16 + 20 = 9\ \mathrm{(D)}$.