is irreducible, it follows that $g^*$ or $h^*$, say $g^*$, is a constant. Since $f^* \ne 0$, it follows that $g^*$ is a non-zero constant. Therefore $$4 = \deg (f^*) = \deg (g^*) + \deg (h^*)= 0 + \deg (h^*).$$ Hence $\deg (h^*) = 4$. Since $\deg (h) \ge \deg (h^*)$, it follows that $\deg (h) = 4$. Therefore $4 = \deg (f) = \deg (g) + \deg (h) = \deg (g) + 4$, and hence $g$ is a constant.

Thus $f$ is irreducible, as desired. $\Box$

\[CM\] Let $k$ be a field with ${\mbox{char~}}k \ne 2$, and let $R=k[x_{ij} : 1 \le i < j \le n+1]$, $n \ge 2$, be the polynomial ring over $k$ in the set $X=\{x_{ij} : 1 \le i < j \le n+1\}$ of $(n+1)n/2$ indeterminates. For $1 \le i \le j \le n+1$, let us make the convention that $x_{i,j} = x_{j,i}$ and $x_{j,j} = 0$. Let $M$ be the Cayley-Menger determinant in $X$, i.e., $M$ is the $(n+2)\times (n+2)$ determinant whose entries $c_{i,j}$, $0 \le i, j \le n+1$, are given by $$\begin{aligned}
 c_{i,j} &=& \left\{ \begin{array}{lcl} 0 &\mbox{if}& i=j,\\  
                                 1 &\mbox{if}& i \ne j  \mbox{~and~} ij=0,\\
 x_{i,j}^2 & \mbox{otherwise.} \end{array}\right.
 \end{aligned}$$ 