# 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
$(x-3)(x^2+x+1)$

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What do you mean by "factor in real numbers"? If you mean "splits into real linear factors", then a lot of them satisfy that condition. – wj32 Oct 31 '12 at 8:28
Oh sorry the question is which of them cannot be factored in real numbers. Hence my answer – Alek Oliver Oct 31 '12 at 8:31
Your choice seems to be wrong - $x^2+x+1$ cannot be factored, but you chose the third polynomial instead of the second polynomial. – wj32 Oct 31 '12 at 8:32
You are right. It was exactly what i meant, just copied wrong polynomial when i had to do the last answer. Thanks – Alek Oliver Oct 31 '12 at 8:39

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 $i^2 = -1$.