3. Suppose $f$ is a function that assigns to each real number $x$ a value $f(x)$, and suppose the equation
$$
f\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)=f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)+f\left(x_{4}\right)+f\left(x_{5}\right)-8
$$
holds for all real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. What is $f(0)$ ?
Solution:
2

Plug in $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$. Then the equation reads $f(0)=5 f(0)-8$, so $4 f(0)=8$, so $f(0)=2$.