We know that the following equation has two pairs of roots whose sum is 0. Solve the equation.

$$
x^{6}-2 x^{5}-9 x^{4}+14 x^{3}+24 x^{2}-20 x-20=0
$$
I. Solution. Let one pair of roots be $\alpha$ and $-\alpha$, the other $\beta$ and $-\beta$ (we can assume that $\alpha, \beta > 0$, since $x=0$ does not satisfy (1)), and the remaining two roots be $\gamma, \delta$. Then the left side of (1) as a product of root factors is:

$$
\begin{gathered}
\left(x^{2}-\alpha^{2}\right)\left(x^{2}-\beta^{2}\right)(x-\gamma)(x-\delta)= \\
=\left[x^{4}-\left(\alpha^{2}+\beta^{2}\right) x^{2}+\alpha^{2} \beta^{2}\right]\left[x^{2}-(\gamma+\delta) x+\gamma \delta\right]
\end{gathered}
$$

and by the equality of the two expressions,

$$
\begin{array}{lr}
\text{coefficient of } x^{5} & -(\gamma+\delta)=-2, \\
\text{coefficient of } x^{4} & \gamma \delta-\left(\alpha^{2}+\beta^{2}\right)=-9, \\
\text{coefficient of } x^{3} & \left(\alpha^{2}+\beta^{2}\right)(\gamma+\delta)=14, \\
\text{coefficient of } x^{2} & -\gamma \delta\left(\alpha^{2}+\beta^{2}\right)+\alpha^{2} \beta^{2}=24, \\
\text{coefficient of } x & -\alpha^{2} \beta^{2}(\gamma+\delta)=-20, \\
\text{coefficient of } x^{0} & \alpha^{2} \beta^{2} \gamma \delta=-20 .
\end{array}
$$

It is cautionary that we have obtained 6 equations for the 4 coefficients appearing in (2).

Dividing (5) and (7) by (3), we get $\alpha^{2}+\beta^{2}=7$ and $\alpha^{2} \beta^{2}=10$, and dividing the latter by (8), we get $\gamma \delta=-2$. With these, the unused (4) and (6) are also satisfied. Proceeding further,

$$
\alpha^{2}, \beta^{2} \text{ have values } 2 \text{ and } 5, \gamma \text{ and } \delta \text{ have values } 1+\sqrt{3} \text{ and } 1-\sqrt{3},
$$

and thus the roots of the equation are:

$$
\sqrt{2},-\sqrt{2}, \sqrt{5},-\sqrt{5}, 1+\sqrt{3}, 1-\sqrt{3}
$$

Artur Csetényi (Kiskunhalas, Szúcs J. Secondary School, 8th grade)

II. Solution. We continue to use the above notation. $\alpha$ and $-\alpha$ satisfy (1), so

$$
\begin{aligned}
& \alpha^{6}-2 \alpha^{5}-9 \alpha^{4}+14 \alpha^{3}+24 \alpha^{2}-20 \alpha-20=0 \\
& \alpha^{6}+2 \alpha^{5}-9 \alpha^{4}-14 \alpha^{3}+24 \alpha^{2}+20 \alpha-20=0
\end{aligned}
$$

Accordingly, $\alpha$ can only be a number that also satisfies the difference of these two equalities:

$$
4 \alpha\left(\alpha^{4}-7 \alpha^{2}+10\right)=0
$$

However, $\alpha$ cannot be 0, as it is not a root of (1). From the resulting

$$
\alpha^{4}-7 \alpha^{2}+10=0
$$

equation, the values of $\alpha^{2}, \beta^{2}$ are 2 and 5 in some order, but the order is irrelevant.

The product of the root factors corresponding to the found roots is $\left(x^{2}-2\right)\left(x^{2}-5\right)=x^{4}-7 x^{2}+10$, and (1) can be divided by this (without remainder), and the quotient is $x^{2}-2 x-2$. This means that $x_{1,2}= \pm \sqrt{2}$ and $x_{3,4}= \pm \sqrt{5}$ are indeed roots of (1), and the other roots are from the equation

$$
x^{2}-2 x-2=0
$$

which gives $x_{5,6}=1 \pm \sqrt{3}$.

László Kóczy (Budapest, Fazekas M. Secondary School, 2nd grade)

III. Solution. Let the pair of roots still be $\alpha$ and $-\alpha$, and as we have seen, $a \neq 0$, so the left side of (1) is divisible by the product of the corresponding root factors

$$
(x-\alpha)(x+\alpha)=x^{2}-\alpha^{2}=x^{2}-a
$$

where $a=\alpha^{2}$. We will determine this. Performing the division, the quotient polynomial $h(x)$ and the remainder polynomial $m(x)$ are:

$$
\begin{aligned}
& h(x)=x^{4}-2 x^{3}+(a-9) x^{2}+(14-2 a) x+\left(24-9 a+a^{2}\right) \\
& m(x)=\left(-20+14 a-2 a^{2}\right) x+\left(-20+24 a-9 a^{2}+a^{3}\right)
\end{aligned}
$$

The latter must be identically 0, so $a$ is a number for which the following must simultaneously hold:

$$
\begin{gathered}
-20+14 a-2 a^{2}=0 \\
20+24 a-9 a^{2}+a^{3}=0
\end{gathered}
$$

The roots of (11) are $a_{1}=2$ and $a_{2}=5$, and both are also roots of (12). This way, we again obtain the roots $x_{1,2}= \pm \sqrt{2}, x_{3,4}= \pm \sqrt{5}$ of (1), and we proceed as in the completion of the II. solution.

Remarks. 1. If we select the pairs of roots one by one from (1), we get the quotient of the first division from (10) by substituting $a=2$ (or $a=5$).

2. The reader should explain the relationship between the left side of (11) and the divisor leading to (9).