# Solution:

Notice that when $x$ is replaced by $3-x$, the expression $3-x$ changes to $x$. That is, the pair $f(x)$ and $f(3-x)$ is invariant under this substitution. Replace $x$ with $3-x$ in the equation given in the problem. We get:

$4 f(x) - f(3-x) = 3(3-x)^2 - 4(3-x) - 3 = 3x^2 - 14x + 12$. Express $f(x)$ from the system: $\left\{\begin{array}{l}4 f(3-x) - f(x) = 3x^2 - 4x - 3 \\ 4 f(x) - f(3-x) = 3x^2 - 14x + 12\end{array}\right.$. Multiply the second equation of the system by 4 and add it to the first, we get: $15 f(x) = 15x^2 - 60x + 45$; $f(x) = x^2 - 4x + 3$. Form the equation: $x^2 - 4x + 3 = 8 ; x^2 - 4x - 5 = 0$; its roots are $x_1 = -1 ; x_2 = 5$. The product of the roots is ( -5 ).
6. 

Find the roots of the equation $f(x)=8$, if $4 f(3-x)-f(x)=3 x^{2}-4 x-3$ for any real value of $x$. In your answer, specify the product of the found roots. #