The [function](https://artofproblemsolving.com/wiki/index.php/Function) $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$. If the equation $f(x)=0$ has exactly four distinct real [roots](https://artofproblemsolving.com/wiki/index.php/Root_(polynomial)), then the sum of these roots is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$
Let one of the roots be $r_1$. Also, define $x$ such that $2+x=r_1$. Thus, we have $f(2+x)=f(r_1)=0$ and $f(2+x)=f(2-x)$. Therefore, we have $f(2-x)=0$, and $2-x$ is also a root. Let this root be $r_2$. The sum $r_1+r_2=2+x+2-x=4$. Similarly, we can let $r_3$ be a root and define $y$ such that $2+y=r_3$, and we will find $2-y$ is also a root, say, $r_4$, so $r_3+r_4=2+y+2-y=4$. Therefore, $r_1+r_2+r_3+r_4=4+4=8, \boxed{\text{E}}$.