Answer:
y = 0, y = -1 (Multiplicity 2) and y = 1 Multiplicity 2)
Step-by-step explanation:
y^5 - 2y³ + y = 0
y(y^4 - 2y^2 + 1) = 0
So one zero is y = 0.
y^4 - 2y^2 + 1 = 0
(y^2 - 1)(y^2 - 1) = 0
(y - 1)((y + 1)(y - 1)(y + 1) = 0
so y = -1 (Multiplicity 2) and y = 1 Multiplicity 2).
