Let's determine the roots of the following equation:

$$
x^{10}-5 x^{8}+4 x^{6}-64 x^{4}+320 x^{2}-256=0
$$
The left side of the equation can be factored, indeed, by factoring out $x^{6}$ from the first 3 terms and -64 from the next 3, the same polynomial appears in both parentheses:

$$
x^{6}\left(x^{4}-5 x^{2}+4\right)-64\left(x^{4}-5 x^{2}+4\right)=\left(x^{6}-64\right)\left(x^{4}-5 x^{2}+4\right)
$$

Thus, the roots of (1) are all those values of $x$ for which at least one of the two factors is 0, that is, for which

$$
\begin{array}{ll}
\text{or} & x^{6}-64=0 \\
\text{or} & x^{4}-5 x^{2}+4=0
\end{array}
$$

(2) can be further factored based on the identity $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$:

$$
x^{6}-64=\left(x^{2}\right)^{3}-4^{3}=\left(x^{2}-4\right)\left(x^{4}+4 x^{2}+16\right)
$$

Thus, (2) is satisfied if

$$
\begin{aligned}
& x^{2}-4=(x-2)(x+2)=0, \quad \text{and if} \\
& x^{4}+4 x^{2}+16=\left(x^{2}+4\right)^{2}-4 x^{2}=\left(x^{2}-2 x+4\right)\left(x^{2}+2 x+4\right)=0
\end{aligned}
$$

(3) can be written as:

$$
\left(x^{4}-4 x^{2}\right)-\left(x^{2}-4\right)=\left(x^{2}-1\right)\left(x^{2}-4\right)=(x+1)(x-1)(x+2)(x-2)=0
$$

Thus, its roots are:

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

(4) does not provide new roots, as it is only satisfied by the found $x_{3}$ and $x_{4}$, and (5) has no (real) roots, because both factors on its left side are positive for all (real) $x$:

$$
\begin{aligned}
& x^{2}-2 x+4=(x-1)^{2}+3>0 \\
& x^{2}+2 x+4=(x+1)^{2}+3>0
\end{aligned}
$$

According to this, the roots of (1) are the numbers under (6).

János Pető (Budapest, Kölcsey F. Gymnasium, 3rd grade)

Remarks. 1. $x_{3}$ and $x_{4}$ are double roots of (1), indeed, based on the repeated factorization, the left side of (1) can be written as:

$$
(x+1)(x-1)(x+2)^{2}(x-2)^{2}\left(x^{2}-2 x+4\right)\left(x^{2}+2 x+4\right)
$$

2. The left side of (2) can also be factored based on the quoted identity as follows:

$$
x^{6}-64=\left(x^{3}-8\right)\left(x^{3}+8\right)=(x-2)\left(x^{2}+2 x+4\right)(x+2)\left(x^{2}-2 x+4\right)
$$