The polynomial $f(x) = x^{4} + ax^{3} + bx^{2} + cx + d$ has real coefficients, and $f(2i) = f(2 + i) = 0.$ What is $a + b + c + d?$
Just like in Solution 1 we realize that the roots come in conjugate pairs. Which means the roots are $2i,i+2,-2i,2-i$ So our polynomial is (1) $f(x) = (x-2i)(x+2i)(x-i-2)(x-2+i)$ Looking at the equation of the polynomial $f(x) = x^4+ax^3+bx^2+cx+d$. We see that $a+b+c+d = f(1)-1$ If we plug in $1$ into equation (1) we get $f(1) = (1-2i)(1+2i)(-1-i)(-1+i)$. Now if we multiply a complex number by its conjugate we get the sum of the squares of its real and imaginary parts. Using this property on the above we multiply  and get $f(1) = (1-2i)(1+2i)(-1-i)(-1+i) = (1^2+2^2)(1^2+1^2) = 10$ So the answer is $f(1) -1 = 10 - 1  = 9$. $\framebox{D}$ $\fbox{9}$.