First, the trivial part is to check for linear factors. If my calculations are right, the only possibility is $(x+1)$ could be a factor when $n \equiv 1 \pmod{2}$. Otherwise, assume all factors to be at least quadratic, and let $W(x) = f(x)g(x)$. Work in $\mathbb{F}_2[x]$. Then $W(x) = x^{n-1}(x-3)$. Since this is the irreducible in our field, we therefore can write (WLOG) $$f(x) = x^i(x-3) + 2h(x), g(x) = x^{n-1-i} + 2k(x)$$ , where $1 \le i \le n-3$(because we already checked all linear factors) and $\deg(h(x)) < \deg(f(x))$ and $\deg(k(x)) < \deg(g(x))$. Now we see that $$f(x)g(x) = x^n-3x^{n-1}+2h(x)x^{n-1-i}+2k(x)x^i(x-3)+4h(x)k(x) = W(x) = x^n-3x^{n-1}+2x^{n-2}+6$$ , so after subtracting and dividing by 2, $$h(x)x^{n-1-i}+k(x)x^i(x-3)+2h(x)k(x) = x^{n-2}+3$$ . But since $1 \le i \le n-3$, we can match constant terms to see that the constant term of $2h(x)k(x)$ is 3. But this is clearly a contradiction, because 3 is odd. So all odd $n$ work, and all evens don't. Remark: What I essentially did is prove Extended Eisenstein's(Theorem 3).

We claim that the answer is all odd integers $n \ge 3.$ For odd $n \ge 3$, $x+1$ divides $W(x)$ and so odd $n \ge 3$ clearly work. Let's now show that even $n \ge 4$ do not work. We will work in $\mathbb{Z}_3 [x].$ Notice that the polynomial $W(x)$ is just $x^{n-2}(x^2+2)$ in $\mathbb{Z}_3[x].$ Assume, for contradiction, that $W(x) = A(x) B(x)$ for some nonconstant $A, B \in \mathbb{Z} [x].$ WLOG assume that $A, B$ are both monic. Notice that, in $\mathbb{Z}_3 [x]$, we have $A B = x^{n-2}(x+1)(x+2).$ We therefore have one of two cases, neglecting the order of $A(x)$ and $B(x).$ Case 1. For some $a+b = n-2$, $A(x) = (x+1)x^a$ and $B(x) = (x+2)x^b$ in $\mathbb{Z}_3[x]$ If neither of $a$ and $b$ is zero, then the constant term of $AB$ is divisible by $9$, clearly absurd. Else, $a$ or $b$ is zero, meaning that a monic linear polynomial divides $W(x).$ This is clearly absurd for $n$ even. Case 2. For some $a+b = n-2$, $A(x) = (x+1)(x+2)x^a$ and $B(x) = x^b$ in $\mathbb{Z}_3 [x].$ Again, one of $a, b$ is zero for similar reasons as before, and so it must be $a$. Clearly, $A(x)$ must then be of the form $x^2 + 3kx + 2$ or $x^2 + 3kx - 1.$ It's easy to see that these don't work if $k = 0$. Also, $x^2 + 3x + 2$ and $x^2 - 3x +2$ easily don't work. Otherwise, it can be checked by the Quadratic Formula that $A(x)$ has a real root of absolute value greater than $3$, say $r$. If $r< 0$, then $r^n - 3r^{n-1} + 2r^{n-2} + 6 > 0$ because each term is positive. If $r > 0$, then $r^n - 3r^{n-1} > 0$ and $2r^{n-2} + 6 > 0$. In either case $W(r) > 0$, contradicting the fact that $r$ is a root of $W(x).$ As we've now exhausted all cases, we can conclude that even $n \ge 4$ do not work. Thus, the answer is all odd $n \ge 3.$ $\square$

Here is my solution for this problem Solution It's easy to see that: $n \equiv 1$ (mod $2$) satisfy the problem With $n = 4$, we have: $W(x) = x^4 - 3x^3 + 2x^2 + 6$ Since: $W(x) > 0, \forall x \in \mathbb{Z}$, then: $W(x)$ can't be written as a product of two non-constant polynomials with integer coefficients and odd degrees So: $W(x) = (x^2 + ax + b)(x^2 + cx + d)$ Hence: $a + c = - 3$, $ac + b + d = 2$, $ad + bc = 0$, $bd = 6$ These equations have no integer solutions Then: $W(x) = x^4 - 3x^3 + 2x^2 + 6$ is irreducible With $n \ge 6$, suppose: $W(x) = Q(x)R(x)$ with $Q(x)$, $R(x)$ are non-constant polynomials with integer coefficients WLOG, we consider: $\text{deg} Q = k \ge \dfrac{n}{2}$ Let $Q(x) = (x - x_1)(x - x_2).....(x - x_k)$ with $x_1, x_2, .....,x_k \in \mathbb{C}$ So: $|Q(0)| = |x_1x_2.....x_k| \in \{1; 2; 3; 6\}$ But: $W(x_1) = W(x_2) = ..... = W(x_k) = 0$ then: $|(Q(0))^nQ(1)Q(2)| = 6^k$ If $|Q(0)| = 1$, we have: $|Q(1)Q(2)| = 6^k \le 36$, which is a contradiction since: $k \ge \dfrac{n}{2} \ge 3$ If $|Q(0)| = 6$, we have: $|(Q(0))^nQ(1)Q(2)| = 6^k$ So: $k = n$, which is a contradiction since $\text{deg} R > 0$ If $|Q(0)| = 2$, we have: $|Q(1)Q(2)| = \dfrac{3^k}{2^{n - k}}$, conflict with $|Q(1)|$, $|Q(2)|$ $\in$ $\mathbb{Z}$ If $|Q(0)| = 3$, we have: $|Q(1)Q(2)| = \dfrac{2^k}{3^{n - k}}$, conflict with $|Q(1)|$, $|Q(2)|$ $\in$ $\mathbb{Z}$ Hence with $n \equiv 0$ (mod $2$), $W(x)$ is irreducible

Thank you, khanhnx! Your solution is nice and accessible...