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

$$
\left\{\begin{array}{l}
a+c=4 \\
a c+b+d=6 \\
a d+b c=5 \\
b d=2
\end{array}\right.
$$
Answer: $(3,2,1,1)$ and $(1,1,3,2)$.

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

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

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

$$
x^{4}+4 x^{3}+6 x^{2}+5 x+2=\left(x^{2}+3 x+2\right)\left(x^{2}+x+1\right)
$$

Since $x^{2}+x+1$ has no real roots, it cannot be represented as a product of two first-degree polynomials, so the identity

$$
\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=\left(x^{2}+3 x+2\right)\left(x^{2}+x+1\right)
$$

is possible only in two cases

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + a x + b = x ^ { 2 } + 3 x + 2 } \\
{ x ^ { 2 } + c x + d = x ^ { 2 } + x + 1 }
\end{array} \text { or } \left\{\begin{array}{l}
x^{2}+a x+b=x^{2}+x+1 \\
x^{2}+c x+d=x^{2}+3 x+2
\end{array}\right.\right.
$$

then in the first case we get $a=3, b=2, c=1, d=1$, and in the second $-a=1, b=1, c=3, d=2$.