Let $\displaystyle z^{5}+2z^{3}+z = p(z).$

By the remainder theorem you know that:

$\displaystyle a$ is a root of $\displaystyle p(z) \Leftrightarrow p(z)=(z-a)q(z)$ for some poloynomial $\displaystyle q(z).$

You did this correctly for $\displaystyle a=0$ to get

$\displaystyle p(z) = (z-0)q(z) = z(z^{4}+2z^{2}+1).$

Then by your substitution you found that $\displaystyle u=z^{2}=-1$, hence $\displaystyle z=(-1)^{\frac{1}{2}}=\pm i$, so that we get

$\displaystyle p(z)=z(z^{2}+1)(z^{2}+1) = z(z+i)(z-i)(z+i)(z-i)$.

Therefore the roots are $\displaystyle 1,i,-i$. The reason that you got 3 distinct roots instead of the 5 you might expect with an order 5 polynomial is that the factors $\displaystyle (z+i)$ and $\displaystyle (z-i)$ are both repeated in the (unique) factorisation of $\displaystyle p(z).$ So really there are 5 roots, it's just that $\displaystyle i$ and $\displaystyle -i$ are repeated.