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I'm given a problem to factorize $$ P(x)=x^5+x+1 $$

I've done the following:

$$ P(x)=(x^5+x^4+x^3)-(x^4+x^3+x^2)+(x^2+x+1)= (x^2+x+1)(x^3-x^2+1)$$

Is it possible to prove that this cannot be factorized any further?

Lazar Ljubenović
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    Assuming you're factorising over $\mathbb Q $: Have you heard of Gauss' Lemma? If so, note that if a quadratic or cubic has a proper factor, then it must have a linear factor. – Matt Feb 07 '12 at 22:40
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    Over the rationals, no. For if you could factor $x^3-x^2+1$ further, one of the factors at least would be linear, and that factor would have a rational zero (root). But the equation $x^3-x^2+1$ has no rational solutions. To see why, look up Rational Root Theorem. It is not hard to understand and prove. Thus the only conceivable candidates for rational root here are $x=1$ and $x=-1$, and neither works. Good work getting the first factorization. – André Nicolas Feb 07 '12 at 22:59

1 Answers1

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It depends what you are factoring over. Notice that if we use the quadratic formula, $x^2 + x + 1$ has roots $-\frac{1}{2} \pm \frac{i\sqrt{3}}{2}$, so it can be factorized as

$$(x^2 + x + 1) = (x - (-\frac{1}{2} + \frac{i\sqrt{3}}{2}))(x - (-\frac{1}{2} - \frac{i\sqrt{3}}{2}))$$

If you are factoring over just the real numbers, finding the roots shows that this polynomial (at least the $x^2+x+1$ portion) cannot be reduced further.

The $x^3-x^2+1$ portion can be factored further over the reals because it has one real root, but it is not further reducible over just the rational numbers.

So the point at which a polynomial has no further factors depends on the set over which you are factoring.

Kurtis Zimmerman
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