Unless the Fundamental Theorem of Algebra is wrong, there had better be three zeros since this is a third degree polynomial. Just because two of them are equal doesn't change the fact that there are three factors any one of which could be zero.
Take out
Then factor the perfect square polynomial that remains:
So the zeros are 0, 1, and 1.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
