# Find a degree-4 polynomial in $\mathbb{Q}[x]$ that is not irreducible but also has no roots.

Let $$F$$ be a field, $$f \in F[x]$$ of degree 2 or 3. Theorem: If $$f$$ has no roots, then $$f$$ is irreducible.

I have shown in several examples the above theorem holds. I am trying to find an example where the theorem does not hold in a degree-4 polynomial in $$\mathbb{Q}[x]$$.

The quartic polynomial that I have tried was $$(x^4 - 22x^2 + 1)$$. By the Rational Roots Theorem, the only possible rational roots for $$(x^4 - 22x^2 + 1)$$ are $$\pm$$1. But $$(1)^4 - 22(1)^2 + 1 \neq 0$$ and $$(-1)^4 - 22(-1)^2 + 1 \neq 0$$. Hence $$(x^4 - 22x^2 + 1)$$ has no roots. As I'm trying to show that $$(x^4 - 22x^2 + 1)$$ is reducible, I know that (from WolfRamAlpha) "A polynomial is said to be irreducible if it cannot be factored into nontrivial polynomials over the same field." So, if I can't factor $$(x^4 - 22x^2 + 1)$$ into nontrivial polynomials over $$\mathbb{Q}[x]$$, then $$(x^4 - 22x^2 + 1)$$ over $$\mathbb{Q}[x]$$ is irreducible. Since I don't know how to factor $$(x^4 - 22x^2 + 1)$$ into nontrivial polynomials over $$\mathbb{Q}[x]$$, can I conclude that $$(x^4 - 22x^2 + 1)$$ over $$\mathbb{Q}[x]$$ is reducible?

• consider $(x^2+1)(x^2+2)$ Dec 3, 2019 at 2:31
• Or more simply, $(x^2+1)^2$. Dec 3, 2019 at 2:38
• In the last two sentences, do you mean that if you can't factor, then it's irreducible, not reducible? Dec 3, 2019 at 2:41
• @GoranMalic Just edited it. Thanks Dec 3, 2019 at 2:42
• The last sentence still says reducible; other than that, it’s pretty much a repeat of the previous sentence Dec 3, 2019 at 2:54

## 1 Answer

You can't conclude that your polynomial is irreducible until you prove that it cannot be factored nontrivially over $$\Bbb Q$$. This will take some effort, presumably.

You could write $$x^4-22x^2+1=(x^2+bx+c)(x^2+dx+e)$$, get a system of equations, and try to solve it.

Namely, $$\begin{cases} b+d=0\\bd+e+c=-22\\cd+eb=0\\ec=1\end{cases}$$.

So, I get that $$(c-e)d=0$$, hence either $$c=e=\pm1$$ or $$b=d=0$$.

But $$c=e=1\implies b=-24/d\implies d-24/d=0\implies d=\sqrt{24}\not\in\Bbb Q$$. Similarly if $$c=e=-1$$.

On the other hand, $$b=d=0\implies e+1/e=-22\implies e^2+22e+1=0$$. This gives again that $$e$$ is not rational, as the discriminant $$480$$ isn't a perfect square.

Alternatively, how about starting from the factorization, as in $$((x-\sqrt2)(x+\sqrt2))^2=(x^2-2)^2=x^4-4x^2+4$$? Unlike the examples in the comments, here the roots are real. Not that it matters.