Do the grouping $2y^{3}+y^{2}-2y-1=\left(2y^{3}+y^{2}\right)+\left(-2y-1\right)$, and factor out $y^{2}$ in the first and $-1$ in the second group.
$$y^{2}\left(2y+1\right)-\left(2y+1\right)$$
Factor out common term $2y+1$ by using distributive property.
$$\left(2y+1\right)\left(y^{2}-1\right)$$
Consider $y^{2}-1$. Rewrite $y^{2}-1$ as $y^{2}-1^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
