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?