$$
x^{3}+a x^{2}+b x+6=0
$$

In the equation, determine $a$ and $b$ such that one root is 2 and another root is 3. What is the third root?
I. solution: By substituting the given roots into the equations, we get

$$
\begin{aligned}
& 4 a+2 b+14=0 \\
& 9 a+3 b+33=0
\end{aligned}
$$

from which

$$
a=-4, b=1
$$

Substituting these values of $a$ and $b$ into our equation, we arrive at

$$
x^{3}-4 x^{2}+x+6=0
$$

On the other hand, if we denote the sought third root by $x_{3}$, the factorized form of our equation is

$$
(x-2)(x-3)\left(x-x_{3}\right)=0
$$

From (1) and (2), it follows that

$$
x-x_{3}=\frac{x^{3}-4 x+x+6}{(x-2)(x-3)}=\frac{x^{3}-4 x^{2}+x+6}{x^{2}-5 x+6}=x+1
$$

thus

$$
x_{3}=-1
$$

Balatoni István (Veszprém, Lovassy László g. II. o. t.)

II. solution: The factorized form of a third-degree equation (if the coefficient of the third-degree term is 1) is

$$
\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)=0
$$

or

$$
x_{3}-\left(x_{1}+x_{2}+x_{3}\right) x_{2}+\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right) x-x_{1} x_{2} x_{3}=0 .
$$

In this case,

$$
\begin{aligned}
& a=-\left(x_{1}+x_{2}+x_{3}\right) \\
& 6=x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3} a \\
& 6=-x_{1} x_{2} x_{3}
\end{aligned}
$$

furthermore,

$$
x_{1}=2 \quad \text { and } \quad x_{2}=3
$$

From (3),

$$
6=-6 x_{3},
$$

from which

$$
x_{3}=-1
$$

and thus from (1),

$$
a=-(2+3-1)=-4
$$

and from (2),

$$
b=6-2-3=1
$$

Vastag György (Bp. XXI., Jedlik Ányos g. II. o. t.)