Problem 5. Solve the system of equations in real numbers:

$$
\left\{\begin{array}{l}
a+c=-7 \\
a c+b+d=18 \\
a d+b c=-22 \\
b d=12
\end{array}\right.
$$
Answer: $(-5,6,-2,2)$ and $(-2,2,-5,6)$

Solution. Let $x^{2}+a x+b$ and $x^{2}+c x+d$ be the two quadratic polynomials, the coefficients of which are the roots of the given system. Then

$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=x^{4}-7 x^{3}+18 x^{2}-22 x+12$.

By trial, we find that the numbers 2 and 3 are roots of this polynomial, from which

$$
x^{4}-7 x^{3}+18 x^{2}-22 x+12=\left(x^{2}-5 x+6\right)\left(x^{2}-2 x+2\right)
$$

Since $x^{2}-2 x+2$ has no real roots, it cannot be represented as a product of two first-degree polynomials, so the identity

$$
\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=\left(x^{2}-5 x+6\right)\left(x^{2}-2 x+2\right)
$$

is possible only in two cases

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + a x + b = x ^ { 2 } - 5 x + 6 } \\
{ x ^ { 2 } + c x + d = x ^ { 2 } - 2 x + 2 }
\end{array} \text { or } \left\{\begin{array}{l}
x^{2}+a x+b=x^{2}-2 x+2 \\
x^{2}+c x+d=x^{2}-5 x+6
\end{array}\right.\right.
$$

then in the first case we get $a=-5, b=6, c=-2, d=2$, and in the second $-a=-2, b=2, c=-5$, $d=6$.