I'm asked to factor the fifth degree polynomial $p(x) = x^5 + 2x^4 + x + 2$ into irreducible factors in $\mathbb{Q}[x]$, $\mathbb{R}[x]$, and $\mathbb{C}[x]$, respectively.
After guessing the first real root $-2$ and dividing $p(x)$ by $(x + 2)$, I get $x^4 + 1$, which has the complex roots $e^{\frac{i\pi}{4}}, e^{-\frac{i\pi}{4}}, e^{\frac{3i\pi}{4}},$ and $e^{-\frac{3i\pi}{4}}$.
The factorisation of $p(x)$ in $\mathbb{C}[x]$ is therefore:
$$p(x) = (x+2)(x-e^{\frac{i\pi}{4}})(x-e^{-\frac{i\pi}{4}})(x-e^{\frac{3i\pi}{4}})(x-e^{-\frac{3i\pi}{4}})$$
However, for $\mathbb{R}[x]$ I'm told to take the respective product of each of the two complex conjugate factor pairs (i.e. $(x - e^{\frac{i\pi}{4}}) \cdot (x - e^{-\frac{i\pi}{4}})$ and $(x - e^{\frac{3i\pi}{4}}) \cdot (x - e^{-\frac{3i\pi}{4}})$) to get the real zeros.
Why does the product of two complex conjugate zeros give me a real zero? I understand that those products are real numbers, I'm just confused as to why these make up real zeros of $p(x)$.
"product of two complex conjugate zeros give me a real zero" this is not true.
set $z=\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}$ then $z$ is a root of $x^4+1$.
$(x-z)(x-\overline z)= x^2-(z+ \overline z)x+z\overline z=x^2-\sqrt2x+1$.
Observe that $z\overline z=1$ is not a root of $x^2-\sqrt2x+1$. Also 1 is not a root of $x^5 + 2x^4 + x + 2$