If a+bi is a root of F, then a-bi is too

[Elzair]

Homework Statement

Given f\left(x\right) \in \mathbb{R}\left[x\right], if a + bi is a root of f\left(x\right), then a - bi is a root of f\left(x\right) as well.

Homework Equations

Factor Theorem: if f\left(x\right) is a polynomial over a ring with identity, then a is a root of f\left(x\right) if and only if \left(x - a\right) is a factor of f\left(x\right).

The Attempt at a Solution

I know I did not do this in the standard way, but I was wondering if it still worked.

If a + bi is a root of f\left(x\right), then f\left(a + bi\right) = 0, and so by the Factor Theorem x - \left(a + bi\right) is a factor of f\left(x\right).

Since \left(a + bi\right) \notin \mathbb{R}, \left(x - \left(a + bi\right)\right) \notin \mathbb{R}}\left[x\right].

Therefore, \exists \left(c + di\right) such that \left(x - \left(a + bi\right)\right)\left(x - \left(c + di\right)\right) \in \mathbb{R}}\left[x\right].

\left(x - \left(a + bi\right)\right)\left(x - \left(c + di\right)\right)
=x^{2} - x\left(c + di\right) - x\left(a + bi\right) + \left(a + bi\right)\left(c + di\right)
=x^{2} - x\left(a + c\right) - x\left(b + d\right)i + \left(ac + \left(ad + bc\right)i - bd\right)
=x^{2} - x\left(a + c\right) - x\left(b + d\right)i + \left(ac + \left(ad + bc\right)i - bd\right)
=x^{2} - x\left(a + c\right) + ac - bd \left( ad + bc - x\left(b + d\right)\right)i

If \left(x^{2} - x\left(a + c\right) + ac - bd + \left( ad + bc - x\left(b + d\right)\right)i\right) \in \mathbb{R}}\left[x\right], ad + bc - xb - xd = 0

Set c = a and d = -b.

x^{2} - x\left(2a\right) + a^{2} - b^{2} + \left( ba - ab - x\left(b - b\right)\right)i
= x^{2} - x\left(2a\right) + a^{2} - b^{2} \in \mathbb{R}\left[x\right]

Therefore, \left(x - \left(a - bi\right)\right) is a factor of f\left(x\right)}\left[x\right], so a - bi is a root of f\left(x\right) in \mathbb{R}\left[x\right].

 

[Dick]

You are way overcomplicating this, and missing an important point. Even if f(x) is quadratic it doesn't need to start as x^2+... A polynomial in R[x] has the form f(x)=a_n*x^n+a_{n-1}*x^(n-1)+...+a1*x+a0, where all of the a's are REAL, right? If f(x)=0 what does taking the complex conjugate tell you?