#### Answer

-1, 0, and $\frac{1}{2}$

#### Work Step by Step

First, factor the polynomial completely:
$\\y=x(2x^2+x-1)
\\y=x(2x-1)(x+1).$
Equate each unique factor to zero, then solve each equation:
$x=0 \text{ or } 2x-1=0 \text{ or } x+1=0
\\x=0 \text{ or } x = \frac{1}{2} \text{ or } x=-1.$
Thus, the zeros are 0, -1, and $\frac{1}{2}$, each with multiplicity one.