Here is my question related to the factorization of polynomials over R. I have 2 questions regarding my highlighting text.
1)The first highlight: I don't understand why if we can prove $q(x)$ has real coefficients, then it will have as much $(x-\overline{\lambda})$ as $(x-\lambda)$.
2)The second highlight: Why $x$ can be in $R$ only ? What if $x\in C$ ?
There are as many $(x-\overline{\lambda})$ as $(x-\lambda)$ because $(x-\overline{\lambda})(x-\lambda)$ will have only real coefficients.
Let $\lambda:=a+bi$, then $\overline{\lambda}=a-bi$, so: $$(x-\overline{\lambda})(x-\lambda)=x^2-\lambda x-\overline{\lambda}x+\overline{\lambda}\lambda=x^2-x(\lambda+\overline{\lambda})+\overline{\lambda}\lambda$$ Now, $\lambda+\overline{\lambda}=2a \in \mathbb{R}$, and $\overline{\lambda}\lambda=a^2+b^2 \in \mathbb{R}$
So if there is no $(x-\overline{\lambda})$ for a $(x-\lambda)$, then the polynom will look like this: $Q(x)=(x-\lambda)P(x)$, where $P(x)$ has only real coefficients. $Q(x)=(x-\lambda)P(x)=(x-a-bi)P(x)=xP(x)-aP(x)-ibP(x)$. Now, $xP(x)$ and $aP(x)$ has real coefficients only, but the $-ibP(x)$ has only imaginary, so $Q(x)$'s coefficients are not real.