The complex numbers are famous for having the property of the fundamental theorem of algebra.
All polynomials have at least one root.
Together with being factor-able, this means we can factor every complex polynomial of degree n into a product of n linear factors. (For monic polynomials we can skip $k$)
$$p(z) = k\cdot \prod_{i=1}^n (z-z_i)$$
For real valued polynomials this is not true. We will there also need to allow polynomial factors of degree 2 to reach some irreducible expression.
So it is clear that there exist two possibilities of numbers : those whose polynomials are guaranteed reducible down to 1 and 2 degrees. But what about 3? Which type of numbers will require us to allow also third-degree polynomial factors to factor "completely"?
Edit > So to clarify, I am looking for a type of number so that for any polynomial with coefficients and variables in this set of numbers which I seek will be guaranteed to be able to factor it into factors of order 1,2 or 3 but never higher and that the case 3 by necessity happens for at least one polynomial.
We can always write :
$$p(x) = \sum_{k=0}^N c_kx^k = K \prod_i (c_{i3}x^3+c_{i2}x^2+c_{i1}x+c_{i0})$$
So that there exists at least one p for which there is at least one i so that $c_{i3} \neq 0$
But for some p, all the $c_{i3}$ and also $c_{i2}$ may be $0$.
Best Answer
I'll follow @TorstenSchoeneberg's line of reasoning. You asked about numbers, so I'll write about a field $k$ with $\mathrm{char } k = 0$.
Now let the degree of (the minimal polynomial of) $\alpha \in \overline{k}$ be $n$ and prove that $k(\alpha) = \overline{k}$. Indeed, for $\beta \in \overline{k}$ we have $k(\alpha, \beta) = k(\gamma)$ for some $\gamma$ by the primitive element theorem, so $|k(\gamma)/k| \le n$ (the dimension as a vector space), but $k(\alpha)$ is contained in $k(\gamma)$, so they are equal and $\gamma, \beta \in k(\alpha)$.
This allows us to use
We conclude that a field of characteristic $0$ that has bounded degree $d$ of irreducible polynomials has $d \le 2$. The same holds without the a priori characteristic assumption (the bounded degree implies $\mathrm{char } k = 0$), but this is more difficult to prove.