Pole-zeros of a real-valued causal FIR system
Note the difference between the zeros at $0.3 \pi$ and at $0.8 \pi$.

The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.

At $\theta = 0.8 \pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.

So your zeros are:


2 zeros at $z = e^{\pm j 0.3 \pi}$
2 double zeros at $z = e^{\pm j 0.8 \pi}$