Problem:
The polynomial f(x)=x4+ax3+bx2+cx+d has real coefficients, and f(2i)= f(2+i)=0. What is a+b+c+d?
Answer Choices:
A. 0
B. 1
C. 4
D. 9
E. 16
Solution:
Because f(x) has real coefficients and 2i and 2+i are zeros, so are their conjugates −2i and 2−i. Therefore
f(x)=(x+2i)(x−2i)(x−(2+i))(x−(2−i))=(x2+4)(x2−4x+5)=x4−4x3+9x2−16x+20.
Hence a+b+c+d=−4+9−16+20=9.
OR
As in the first solution,
f(x)=(x+2i)(x−2i)(x−(2+i))(x−(2−i))
so
a+b+c+d=f(1)−1=(1+2i)(1−2i)(−1−i)(−1+i)−1=(1+4)(1+1)−1=9.
The problems on this page are the property of the MAA's American Mathematics Competitions