Right. For the last one, $\displaystyle r^4$ is idempotent and therefore central. Now

$\displaystyle (r+1)^5\,=\,r+1$

$\displaystyle \implies\ 2r^3+2r^2+r\,=\,-r^4$ is central

and

$\displaystyle (r-1)^5\,=\,r-1$

$\displaystyle \implies\ 2r^3-2r^2+r\,=\,r^4$ is central

and so $\displaystyle (2r^3+2r^2+r)-(2r^3-2r^2+r)=4r^2$ and $\displaystyle (2r^3+2r^2+r)+(2r^3-2r^2+r)=4r^3+2r$ are central. Hence $\displaystyle 8(4r^2)=2^5r^2=2r^2$ is central. Moreover

$\displaystyle 4r^3+6r^2+4r\,=\,(r+1)^4-r^4-1$

is central. Hence $\displaystyle (4r^3+6r^2+4r)-(4r^3+2r)-3(2r^2)=2r$ is central. Hence $\displaystyle (4r^3+2r)-2r=4r^3$ is central and so $\displaystyle 8(4r^3)=2^5r^3=2r^3$ is central. Hence $\displaystyle (2r^3+2r^2+r)-2r^3-2r^2=r$ is central.