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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.

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    You are right. It was exactly what i meant, just copied wrong polynomial when i had to do the last answer. Thanks2012-10-31

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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,

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    I think @Patrick would do the trick, but I'm not quite sure. Anyway that's how I tag, see how I tagged you. Either way I got the notice because it's a comment on my answer, so it's okay.2012-11-01