The compound shown is 3-methylpent-1-ene drawn without stereochemistry. The total number of stereoisomers (geometrical and optical together) is

The terminal double bond has two $\mathrm{H}$ on C1, so there is no geometrical isomerism. C3 carries $\mathrm{H}$, $\mathrm{CH}{}_3$, $\mathrm{C}{}_2\mathrm{H}{}_5$ and vinyl, so it is the only stereocentre. The total is $2^1=2$ stereoisomers, the $(R)$ and $(S)$ enantiomers.