Solve the equation

$$
x(x+2)(x+3)(x+5)+8=0
$$
I. Solution: We can observe that the arithmetic mean of the first and fourth factors of the product is equal to the arithmetic mean of the second and third factors, which is $x+\frac{5}{2}$. Let us consider this as a new unknown, so let

$$
x+\frac{5}{2}=z, \quad \text { that is } x=z-\frac{5}{2}
$$

thus we have

$$
\left(z-\frac{5}{2}\right)\left(z-\frac{1}{2}\right)\left(z+\frac{1}{2}\right)\left(z+\frac{5}{2}\right)+8=\left(z^{2}-\frac{25}{4}\right)\left(z^{2}-\frac{1}{4}\right)+8=0
$$

which simplifies to

$$
z^{4}-\frac{13}{2} z^{2}+\frac{153}{16}=0
$$

Solving for the roots and then finding the corresponding $x$ values according to (2):

$$
\begin{aligned}
z_{1} & =\frac{\sqrt{17}}{2}, & x_{1}=\frac{\sqrt{17}-5}{2} \\
z^{2}=\frac{13}{4} \pm 1 & & x_{2}=\frac{-\sqrt{17}-5}{2} \\
z_{2} & =-\frac{\sqrt{17}}{2}, & \\
z_{3} & =\frac{3}{2}, & x_{3}=-1 \\
z_{4} & =-\frac{3}{2}, & x_{4}=-4
\end{aligned}
$$

Thus, we have solved the equation. We see that all four roots are real.

II. Solution: Notice that in the first term of (1), the product of the first and fourth factors differs from the product of the second and third factors only by a constant. Therefore, our equation can be transformed into the following form:

$$
\left(x^{2}+5 x\right)\left(x^{2}+5 x+6\right)+8=0
$$

Let us take the new first factor as a new unknown:

$$
x^{2}+5 x=w
$$

thus we arrive at the equation

$$
w(w+6)+8=0
$$

whose roots are:

$$
w_{1}=-2, \quad w_{2}=-4
$$

Substituting these back into (3), we find that the two $x$ roots corresponding to $w_{1}$ are precisely the $x_{1}$ and $x_{2}$ obtained in the first solution, while the two roots corresponding to $w_{2}$ match the above $x_{3}$ and $x_{4}$.

Remarks. 1. Every equation of the form

$$
(x+a)(x+b)(x+c)(x+d)+e=0
$$

can be solved using either the substitution method applied in the first or the second solution, provided that the numbers $a, b, c, d$ can be paired such that their sums are equal, for example,

$$
a+b=c+d
$$

2. The given equation is simple enough that the integer roots can be found by some trial and error. However, the assumption that the roots are integers is generally unjustified, since, for example, the equation

$$
x(x+2)(x+3)(x+5)+10=0
$$

has no integer roots, and in fact, none of its roots are even real, which can be easily verified by calculating the roots using the substitution method applied in either the first or the second solution. At any rate, it is clear from the product in (1) that $x$ cannot be a (rational) fraction, because if $x=p / q$—where $p$ and $q$ are relatively prime integers, $q>1$—then all four factors are fractions with denominator $q$, and the product of fractions with the same denominator that cannot be simplified cannot be an integer.