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

$$
\left\{\begin{array}{l}
a+c=-1 \\
a c+b+d=-1 \\
a d+b c=-5 \\
b d=6
\end{array}\right.
$$
Answer: $(-3,2,2,3)$ and $(2,3,-3,2)$.

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

$$
\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=x^{4}-x^{3}-x^{2}-5 x+6
$$

By trial, we find that the numbers 1 and 2 are roots of this polynomial, from which

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

Since $x^{2}+2 x+3$ has no real roots, it cannot be represented as a product of two first-degree polynomials, so the identity

$$
\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=\left(x^{2}-3 x+2\right)\left(x^{2}+2 x+3\right)
$$

is possible only in two cases

$$
\left\{\begin{array} { r l } 
{ x ^ { 2 } + a x + b } & { = x ^ { 2 } - 3 x + 2 } \\
{ x ^ { 2 } + c x + d } & { = x ^ { 2 } + 2 x + 3 }
\end{array} \text { or } \left\{\begin{array}{l}
x^{2}+a x+b=x^{2}+2 x+3 \\
x^{2}+c x+d=x^{2}-3 x+2
\end{array}\right.\right.
$$

then in the first case we get $a=-3, b=2, c=2, d=3$, and in the second $-a=2, b=3, c=-3, d=2$.