Let's determine $m$ such that the expression

$$
x^{4}+2 x^{3}-23 x^{2}+12 x+m
$$

is identically equal to the product of the trinomials $x^{2}+a x+c$ and $x^{2}+b x+c$. Determine $a$, $b$, and $c$, as well as the values of $x$ for which the given polynomial is equal to 0.
When we multiply the given trinomials, we get:

$$
x^{4}+(a+b) x^{3}+(2 c+a b) x^{2}+(a c+b c) x+c^{2}
$$

For this expression to be identically equal to the given expression, it is necessary and sufficient that

$$
\begin{gathered}
a+b=2 \\
2 c+a b=-23 \\
c(a+b)=12 \\
c^{2}=m
\end{gathered}
$$

From these equations, we find that

$$
c=6, m=36, a_{1}=-5, b_{1}=7, a_{2}=7, b_{2}=-5
$$

Thus, the given expression can be written as:

$$
x^{4}+2 x^{3}-23 x^{2}+12 x+36=\left(x^{2}-5 x+6\right)\left(x^{2}+7 x+6\right)
$$

This expression equals 0 if

$$
x^{2}-5 x+6=0, \quad \text { or } \quad x^{2}+7 x+6=0
$$

from which we get

$$
x_{1}=2, \quad x_{2}=3, \quad x_{3}=-1, \quad x_{4}=-6
$$

(Kornél Miklóssy, Arad.)

The problem was also solved by: Ádámffy E., Bánó L., Chambrée M., Csada I., Dömény I., Erdős V., Fekete M., Fodor H., Földes R., Fuchs I., Füstös P., Glück J., Haar A., Harsányi Z., Heimlich P., Jánosy Gy., Kiss J., Krampera Gy., Messer P., Kürti I., Paunz A., Rosenberg J., Ruvald S., Sárközy P., Schöffer I., Schuster Gy., Schwarz Gy., Schwarz O., Sonnenfeld J., Strohl J., Wáhl V., Werner M., Mathematikai kör, Bpest, V. ker. fg.