Problem 5. (Option 1). Find all values of the parameters $a, b$, and $c$, for which the set of real roots of the equation $x^{5}+4 x^{4}+a x^{2}=b x+4 c$ consists exactly of the two numbers 2 and -2.
Solution: Since the numbers 2 and -2 are roots of the equation, then

$\left\{\begin{array}{c}32+64+4 a=2 b+4 c \\ -32+64+4 a=-2 b+4 c\end{array} \Leftrightarrow\left\{\begin{array}{c}a=c-16 \\ b=16\end{array}\right.\right.$, substitute into the original equation:

$x^{5}+4 x^{4}+(c-16) x^{2}-16 x-4 c=0$

$x\left(x^{4}-16\right)+4 x^{2}\left(x^{2}-4\right)+c\left(x^{2}-4\right)=0$

$(x-2)(x+2)\left(x^{3}+4 x^{2}+4 x+c\right)=0$

The equation $x^{3}+4 x^{2}+4 x+c=0$ (1) has at least one root. Determine the values of the parameter $c$ for which the roots of the equation are the numbers 2 and -2.

$$
x=2 \text { when } c=-32, \quad x=-2 \text { when } c=0 .
$$

Check if the equation (1) has other roots for these values of $c$:

$$
\begin{array}{ll}
c=-32 & c=0 \\
x^{3}+4 x^{2}+4 x-32=0 & x^{3}+4 x^{2}+4 x=0 \\
(x-2)\left(x^{2}+6 x+16\right)=0 & x\left(x^{2}+4 x+4\right)=0 \\
x=2 & x=-2 ; 0
\end{array}
$$

Therefore, $c=0$ is an extraneous value.

The answer is: $c=-32, b=16, a=-48$.

Answer: $a=-48, b=16, c=-32$.