f(x) = x³ + 2x² + x
f(x) = x.(x² + 2x + 1)
f(x) = x.(x + 1)²
So, f(x) = 0 when x = 0 or when x = -1
But
f(x) = x.(x + 1)² = x.(x + 1).(x + 1)
So we have three roots, x = 0, x = -1 and x = -1, although two of them are equal.
So we have a multiple zero x = -1 with a multiplicity 2.
Alterativa A.