text: how you do this?  [ (1/8)x^2  (2-x)^1/2 ]^2?   [(1/8)*x^2(2-x)^½]^2=?  you need these rules for calculations with potens expression  a^n*b^n=(a*b)^n......@ a^n+a^m=a^(m+n)...@@ a^n^m=a^(n*m)........@@@ (a/b)^n= a^n/b^n......@@@@  now to the work:  according to @ [ (1/8)x^2 (2-x)^1/2 ]^2= ((1/8)x^2)^2*((2-x)^½)^2=  according to @@@ ((1/8)x^2)^2*((2-x)^½)^2= (1/8)^2*x^4)*(2-x)^1)= (1/64)*x^4*(2-x) ============  you see its have the roots x= 2 or x= 0 ==============================