# Factorise $x^6+1$ in rational field vs real field

Recently our class has gone through complex numbers, but I’m still confused about the differences between the rational field $\Bbb Q$ and the real field $\Bbb R$. There were some questions where we have to factorise equations over these fields.

For instance, “factorise $x^6+1$.” (I feel like many people here will find this easy, but I've just started this topic two days ago...)

I could factorise it into $\left(x^2+1\right)\left(x^4-x^2+1\right)$, but does this belong to the real or rational field? And how do I factorise it in the remaining field?

• Welcome to MSE! Here is the MathJax tutorial. I’ll make some edits for you so that you can see how it’s done. – gen-z ready to perish Nov 4 '17 at 2:34
• One broad hint: over $\mathbb{R}$, any real polynomial can be factored into polynomials of degree at most quadratic - can you see why? – Steven Stadnicki Nov 4 '17 at 2:41
• Your factorization works in both the real and rational fields because all the coefficients are rational. As all rationals are real, every factorization in $\Bbb Q$ is a factorization in $\Bbb R$. – Ross Millikan Nov 4 '17 at 2:44
• @StevenStadnicki We haven't learnt super in depth yet. The only idea I could grasp after reading the responses is that in $\Bbb R$ it has rational and irrational whereas $\Bbb Q$ has rational numbers only. (I hope this is right) – Caecilius est in horto Nov 4 '17 at 5:42
• @RossMillikan Right; I'll keep this in mind as I power through! Thank you! – Caecilius est in horto Nov 4 '17 at 5:42

So far, your have factorized $x^6 + 1$ in both the rationals and the reals, though you haven't factorized it into irreducible polynomials. This are the ones that cannot be expressed as the product of two other non-constant polynomials with coefficients in the field. Since $x^2 + 1$ has roots $-i$ and $i$, it's factorization in $\mathbb{C}$ is $(x-i)(x+i)$. These are not polynomials with real or rational coefficients, so $x^2 +1$ is in fact irreducible in $\mathbb{Q}$ and $\mathbb{R}$. As for the other factor, you can check that:

$$x^4-x^2+1=(x^2+1)^2-3x^2=(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)$$

(If this feels like cheating, set $w = x^2$, solve a quadratic equation, and then "go back" to $x^2$). By calculating the discriminant, you can see that these two quadratic polynomials have no real roots and therefore cannot be further factorized in $\mathbb{R}$. This shows that the second factor can be further factorized in $\mathbb{R}$ but not in the rationals, since it decomposes into polynomials that don't have all rational coefficients. We're left with the following decompositions:

$(x^2+1)(x^4-x^2+1)$, in $\mathbb{Q}$

$(x^2+1)(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)$, in $\mathbb{R}$

• Thank you! I was wondering, since the second factor will be factorised in the real field not in rationals, how did you know to make it into(x^2+1)^2-3x^2 ? – Caecilius est in horto Nov 4 '17 at 3:15
• In general $(a+1)^2 = a^2 + 2a + 1$. I exploited that fact with $a = x ^2$, by summing and substracting $3x^2$. This method, if I recall correctly, is called completing the square. There is no actual algorithm for this, sometimes you just "see it" and sometimes you don't. As I've said in the post, if you don't see it, do the substitution, factorize the quadratic into two parts, and then go back. This will let you factorize the polynomial in $\mathbb{C}$ and from there you can recover the two quadratic ones in $\mathbb{R}$. – Guido Nov 4 '17 at 3:19
• Right. Thank you! Subbing $w$ helped! Herding them in brackets was so much easier... – Caecilius est in horto Nov 4 '17 at 5:38

In the real field $x^4-x^2+1$ is further factorizable into

$$x^4-x^2+1=(x^2+1)^2-3x^2=(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1).$$

In the rational field $x^4-x^2+1$ is not factorizable. In the complex field, both $x^2\pm\sqrt{3}x+1$ and $x^2+1$ are factorizable into first-degree polynomials.