As I wrote in my comment, it is more easy to count the reducible degree two polynomials. Such a polynomial is a product of the monic dregree one polynomials. There are three such ($x$, $x-1$ and $x-2$). There are six different products of two of them, hence there are six recucible degree two polynomials. As there are nine monic degree two polynomials alltogether, there are three irreducible ones.
To count the reducible monic degree four polynomials without zero, we argue along the same lines: Such a polynomial is a product $p(x)q(x)$ where $p$, $q$ are monic irreducible degree two polynomials. As we found above, there are three such, we get six reducible, monic, degree four polynomials.
I think I got the proof that no such real polynomial with degree $ \geq 6$ exists.
Let $n \geq 6$
Suppose for contradiction that $z_1,\ldots,z_n \in \mathbb R-\{0\}^n$ are such that $(X-z_1)...(X-z_n)=X^n+\sum_{k=1}^{n-1}z_iX^{n-i}$
Then three useful identities appear $$\sum_{k=1}^{n}z_k=-z_1 \; \; \; \;(1)$$
$$\sum_{\large1\leq i<j \leq n}z_iz_j=z_2 \; \; \; \;(2)$$
$$\prod_{k=1}^n z_k=(-1)^n z_n \; \; \; \;(3)$$
Since $$(\sum_{k=1}^{n}z_k)^2=\sum_{k=1}^{n}z_k^2+2\sum_{\large1\leq i<j \leq n}z_iz_j$$it follows that$$z_1^2=2z_2+\sum_{k=1}^{n}z_k^2$$
Hence $$0< \sum_{k=2}^{n}z_k^2=-2z_2 \; \; \; \;(4)$$ and $$0<\sum_{k=3}^{n}z_k^2=1-(z_2+1)^2 \; \; \; \;(5) $$
$(4)$ and $(5)$ imply $$\; \; \; \;-2<z_2<0 \; \; \; \;(6)$$
thus $(6)$ and $(4)$ imply $$0<\sum_{k=2}^{n}z_k^2 < 4 \; \; \; \; (7)$$
Also $(6)$ and $(5)$ imply $$0<\sum_{k=3}^{n}z_k^2 \leq 1 \; \; \; \; (8)$$
By AM-GM, $$\left(|z_3|^2\ldots|z_{n-1}|^2 \right)^{1/(n-3)} \leq \frac{1}{n-3}\sum_{k=3}^{n-1}z_k^2 \leq \frac{1}{n-3}\sum_{k=3}^{n}z_k^2$$
Hence
$$|z_3|^2\ldots|z_{n-1}|^2 \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{n-3} $$
Squaring, $$|z_3|\ldots|z_{n-1}| \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{\large \frac{n-3}{2}} \leq_{ \large (8)} \dfrac{1}{{(n-3)}^{(n-3)/2}} \; \; \; \; (9)$$
By triangle inequality $(1)$, and Cauchy-Schwarz
$$2|z_1| \leq \sum_{k=2}^{n}|z_k| \leq \sqrt{n-1} \sqrt{\sum_{k=2}^{n}z_k^2} $$
Hence by $(7)$,
$$|z_1| \leq \sqrt{n-1} \; \; \; \; (10)$$
Rewriting $(6)$ as $$|z_2|\lt2 \; \; \; \; (11) $$
Recalling $(3)$ (with $z_n$ cancelled from both sides) and putting together $(9)$, $(10)$ and $(11)$, we have
$$1=|z_1||z_2||z_3|\cdots|z_{n-1}| < \dfrac{ 2\sqrt{n-1}}{{(n-3)}^{(n-3)/2}}$$
This inequality fails for $n\geq 6$.
Contradiction.
I can't prove anything for $n=5$ so maybe the conjecture doesn't hold.
Best Answer
If you start with the polynomial $a=b=0$, then the polynomial will always be $x^2$ and so the process goes on forever. However, in all other cases, it will terminate in finitely many steps.
First, suppose $b=0$ and $a\neq 0$. If $a<0$, the next polynomial is $x^2-a$ which does not have real roots. If $a>0$, the next polynomial is $x^2-ax$ and then the next polynomial is $x^2+a$ which does not have real roots.
Now suppose $b\neq0$. Note then that the initial polynomial does not have $0$ as a root, and consequently neither will any of the others (since the constant coefficient is never $0$). Let us suppose the process goes on forever and say the polynomials are $x^2+ax+b$, $x^2+p_0x+q_0$, $x^2+p_1x+q_1$, $x^2+p_2x+q_2$, and so on.
Note first that the signs of the coefficients $(p,q)$ go in a cycle $$(-,-)\to(-,+)\to (+,+)\to (-,-).$$ For instance, if $p_n$ and $q_n$ are both negative, then the roots of $x^2+p_nx+q_n$ have opposite sign (since their product $q_n$ is negative), so $p_{n+1}$ is negative and $q_{n+1}$ is positive.
Let us now analyze how the absolute value of the linear coefficient changes in this cycle. If $p_n$ and $q_n$ are negative, then $p_{n+1}<0<q_{n+1}$ and $p_{n+1}+q_{n+1}=-p_n>0$, so $|p_{n+1}|<|q_{n+1}|$. Thus $$|p_{n+1}|<\sqrt{|p_{n+1}q_{n+1}|}=\sqrt{|q_n|}\leq\sqrt{|p_n|}.$$ Now suppose $p_n$ is negative and $q_n$ is positive. Then $0<p_{n+1}\leq q_{n+1}$ and so $$|p_{n+1}|=p_{n+1}\leq \frac{p_{n+1}+q_{n+1}}{2}=\frac{-p_n}{2}=\frac{|p_n|}{2}.$$ Finally, suppose $p_n$ and $q_n$ are positive. Then $p_{n+1}\leq q_{n+1}<0$ so $$|p_{n+1}|<|p_{n+1}+q_{n+1}|=|-p_n|=|p_n|.$$
We thus see that in all cases, $|p_{n+1}|\leq\max(|p_n|,1)$. Moreover, in every third step, $|p_{n+1}|\leq\frac{|p_n|}{2}$. It follows that for all sufficiently large $n$, $|p_n|\leq 1$.
Let us now pick $n$ such that $|p_n|\leq 1$ and $p_n$ and $q_n$ are positive. We then have $$p_n^2-4q_n\leq p_n^2-4p_n<0.$$ Thus the discriminant of $x^2+p_nx+q_n$ is negative and it does not have real roots. This is a contradiction.