<s>[INST] How to factor a polynomial in real numbers? Which of the following polynomials can't be factored in real numbers?
Multiple choice question from an old test.
Got the following polynomials:
$x^8$
$(x-3)(x^2+x+1)$
$(x-2)^3(x^2-1)$
$x^8(x^2-1)$
$(x^2-1)^3$
$x(x-1)(x+1)$
Solution:
Solving $x^2+x+1$ isnt possible with real numbers hence D is negative?
There for answer must be $(x-3)(x^2+x+1)$
Edited right answer.
 [/INST] The only polynomial in that list that doesn't split into linear factors is the one with $x^2 + x + 1$ because its discriminant is $-3$. For the other ones, 
\begin{gather}
\begin{aligned}
x^8 & = x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \\
(x-2)^3 (x^2-1) & = (x-2)(x-2)(x-2)(x-1)(x+1) \\
(x^2-1)^3 &= (x-1)(x+1)(x-1)(x+1)(x-1)(x+1) \\
\end{aligned}
\end{gather}
and so on. By the way, the answer $(x-3)(x^2+x+1)$ cannot be factored over the reals, but it can be factored over $\mathbb C$, because the complex numbers have this property that every polynomial splits. As an example, 
$$
x^2 + x + 1 = \left( x - \frac{-1 + \sqrt 3 i}2 \right) \left( x - \frac{-1 - \sqrt 3 i}2 \right),
$$
where $i$ is defined so that $i^2 = -1$. 
Hope that helps,
</s>