and (\[fgh\]) that if $f$ is reducible, then $f^*$ is reducible. It also follows that $$\begin{aligned}
\label{hhh}
\mbox{every factor of a homogeneous polynomial is  homogeneous};\end{aligned}$$ see [@Walker Theorem 10.5, p. 28]. To see that this is not true if $A$ was not an integral domain, consider the factorization $x^2 = (x-2)(x+2)$ in ${\mathbb{Z}}_4[x]$. We end this section by proving a simple property of quadratic polynomials that we shall use later.

\[Q\] Let $A$ be an integral domain with ${\mbox{char~}}A \ne 2$, and let $A[x]$ be the polynomial ring over $A$ in the indeterminate $x$. Let $H =ax^2+bx+c \in A[x]$, with $a \ne 0$, and let $\Delta = b^2 - 4 ac$ be its discriminant. Then

-   If $H$ is reducible, then $\Delta$ is a square (in $A$).

-   If $\Delta$ is a square, and if $2a$ is a unit, then $H$ is reducible.

-   If $H$ is a square, then $\Delta = 0$.

-   If $\Delta = 0$, and $a$ is a square and $2a$ is a unit, then $H$ is a square.

[*Proof.*]{} To prove (i), suppose that $H$ is reducible. Then $H=(U x + V)(ux+v)$ for some $U, V, u, v \in A$. Multiplying out and equating coefficients, 