Question 1022758
Okay, from
y^4-2y^3+y^2=0
let's factor out the y^2 first...we get
y^2(y^2 - 2y + 1) = 0
now factor what's left...
y^2(y - 1)^2 = 0
which gives four solutions, but there are double roots...
y = 0 and y = 1