# Thread: Complex Roots of Polynomial

1. ## Complex Roots of Polynomial

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)

Also
P(-x) = P(x)
therefore P(x) is even

Hence
-(sqrt3 + i) and -(sqrt3 - i) are also roots

from there i got e = 16 (product of roots)
c= ...

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?

2. I agree with your reasoning. I double-checked the book's coefficients in Mathematica. At least one of the roots Mathematica found for the book's coefficients is real!?! But I totally agree that you can find all the roots using conjugates and the function's evenness. So, I would ignore the book's answer.

Cheers!

3. P(*) is a real coefficients polynomial that contains only the even powers of x, so that if $\displaystyle \sigma + i\ \omega$ is a root of P(*), $\displaystyle \sigma - i\ \omega$ , $\displaystyle - \sigma + i\ \omega$ and $\displaystyle -\sigma - i\ \omega$ are also roots od P(*) and is...

$\displaystyle P(x) = (x - \sigma - i\ \omega)\ (x - \sigma + i\ \omega)\ (x + \sigma - i\ \omega)\ (x + \sigma + i\ \omega)$ (1)

Now what You have to do is setting in (1) $\displaystyle \sigma = \sqrt{3}$ and $\displaystyle \omega = 1$ and imposing that...

$\displaystyle P(x) = x^{4} + c\ x^{2} + e$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$