The given compound has two double bonds and the presence of different terminal groups.

The number of geometrical isomers $=2^{\mathrm{n}}$

where $\mathrm{n}=$number of double bonds.

Therefore total geometrical isomers $=2^2=4$

Hence has four geometrical isomers, i.e.,$\left(\text{E,E}\right),\left(\text{E,Z}\right),\left(\text{Z,E}\right)$ and $\left(\text{Z,Z}\right)$.

The compound has also one chiral carbon atom. Therefore, each of the four geometrical isomers will exist as a pair of enantiomers. Thus, in all there are eight stereoisomers.