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

\[
\begin{aligned}
& \left\{\begin{array} { c } 
{ 3 \cdot 2 + 6 \cdot 4 + 2a = 4b + 4c } \\
{ -3 \cdot 2 + 6 \cdot 4 - 2a = 4b + 4c }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
b=16-c \\
a=-16
\end{array},\right.\right. \text { substitute into the original equation: } \\
& x^{5}+4 x^{4}+(c-16) x^{2}-16 x-4 c=0 \\
& x\left(x^{4}-16\right)+4 x^{2}\left(x^{2}-4\right)+c\left(x^{2}-4\right)=0 \\
& (x-2)(x+2)\left(x^{3}+4 x^{2}+4 x+c\right)=0
\end{aligned}
\]

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.

For \(x=2\), \(c=-32\); for \(x=-2\), \(c=0\).

Let's check if the equation (1) has other roots for these values of \(c\):

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

Therefore, \(c=0\) is an extraneous value.

The answer is: \(c=-32, a=-16, b=48\).

Answer: \(a=-16, b=48, c=-32\).