5. Find all solutions of the equation $(x+1)(x+2)(x+3)(x+4)=120$ in the set of complex

(8) numbers.
Solution. By clever grouping, multiplication, and substitution \( t=x^{2}+5 x \), we get:

$$
\begin{aligned}
(x+1)(x+4) \cdot(x+2)(x+3) & =120 \\
\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right) & =120 \\
(t+4)(t+6) & =120 \\
t^{2}+10 t-96 & =0 \\
t_{1}=6, & t_{2}=-16
\end{aligned}
$$

Now we need to solve two quadratic equations:

$$
\begin{aligned}
x^{2}+5 x=6, & & x^{2}+5 x=-16 \\
x^{2}+5 x-6=0, & & x^{2}+5 x+16=0
\end{aligned}
$$

Their solutions are \( x_{1}=-6, x_{2}=1, x_{3}=\frac{-5-i \sqrt{39}}{2}, x_{4}=\frac{-5+i \sqrt{39}}{2} \).

(full solution 8 points)

(solutions \( t_{1,2} \) of equation (*) 4 points)

(other substitutions are possible, e.g., \( t=x^{2}+5 x+5 \))