Problem:
Let p(x)=x2+bx+c, where b and c are integers. If p(x) is a factor of both x4+6x2+25 and 3x4+4x2+28x+5, what is p(1)?
Answer Choices:
A. 0
B. 1
C. 2
D. 4
E. 8
Solution:
Because p(x) is a factor of x4+6x2+25 and of 3x4+4x2+28x+5, it is also a factor of 3(x4+6x2+25)−(3x4+4x2+28x+5), which equals 14x2−28x+70, or 14(x2−2x+5). Therefore, p(x)=x2−2x+5, and p(1)=4.
OR
The unique complete factorization of x4+6x2+25 is a difference of squares:
x4+10x2+25−4x2=(x2+5+2x)(x2+5−2x)
Thus p(x) is either the first quadratic or the second. Long division shows that only the second quadratic is a factor of 3x4+4x2+28x+5. Thus p(x) is the second quadratic.
Note. It was not necessary to assume that b and c are integers. Even if they are allowed to be arbitrary complex numbers, p(x) must still be x2−2x+5. The first solution shows this. (Why?) The second does not. (Why?) Can the second solution be modified so that it does show this?