jmc

If $x^2+bx+b$ is a factor of $x^3+2x^2+2x+c,$ then the other factor must be of the form $x+r.$ Thus, $$(x^2+bx+b)(x+r)=x^3+2x^2+2x+c.$$Expanding, we get $$x^3+(b+r)x^2+(b+br)x+br=x^3+2x^2+2x+c.$$Matching coefficients, we get $$\begin{array}{rl}b+r& =2,\\ b+br& =2,\\ br& =c.\end{array}$$From the equation $b+r=2,$ $r=2-b.$ Substituting into $b+br=2,$ we get $$b+b(2-b)=2.$$Then $b^2-3b+2=0,$ which factors as $(b-1)(b-2)=0.$ Hence, the possible values of $b$ are $\boxed{1,2}.$