P(x) = x^4 + cx^2 + e, where c and e are real
If sqrt3 + i is a root, determine c and e, then factor into quadratic factors with linear factors
Here's my attempt:
Since c and e are real,
then sqrt3 - i is a root (complex conjugates theorem)
P(-x) = P(x)
therefore P(x) is even
-(sqrt3 + i) and -(sqrt3 - i) are also roots
from there i got e = 16 (product of roots)
However the answers have it differently.
It subbed sqrt3 + i and sqrt3 - i into P(x), then solved simultaneously acquiring:
e= - 8, c = 8
Can someone explain the flaw in my reasoning? Or perhaps answers is wrong?