Let's determine all the numbers $x$ for which the following equation holds:

$$
|x+1| \cdot|x-2| \cdot|x+3| \cdot|x-4|=|x-1| \cdot|x+2| \cdot|x-3| \cdot|x+4|
$$
The absolute value of a product is equal to the product of the absolute values of its factors, and vice versa. Therefore, the absolute values of the products

$$
\begin{aligned}
& (x+1)(x-2)(x+3)(x-4)=x^{4}-2 x^{3}-13 x^{2}+14 x+24 \text { and } \\
& (x-1)(x+2)(x-3)(x+4)=x^{4}+2 x^{3}-13 x^{2}-14 x+24
\end{aligned}
$$

are equal. This can happen in two ways: the two products are equal, and thus their difference is 0, or they are negatives of each other, so their sum is 0.

In the first case, rounding the roots to two decimal places,

$$
4 x^{3}-28 x=4 x\left(x^{2}-7\right)=0, x_{1}=0, \quad x_{2,3}= \pm \sqrt{7}= \pm 2.65
$$

In the second case,

$$
\begin{aligned}
& 2 x^{4}-26 x^{2}+48=0, \quad x_{4,5}= \pm \sqrt{6.5+\sqrt{18.25}}= \pm 3.28 \\
& x_{6,7}= \pm \sqrt{6.5-\sqrt{18.25}}= \pm 1.49
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_108487417864fa44c950g-1.jpg?height=667&width=1136&top_left_y=840&top_left_x=489)

Since the right-hand side of (1) is obtained from the left-hand side by replacing \( x \) with \( -x \) - for example, in the first factor \( |-x+1|=|-(x-1)|=|x-1| \), if we have already plotted the left-hand side for the purpose of a graphical solution, the graph of the right-hand side will be the mirror image of the left-hand side across the \( y \)-axis.