The compound shown is hepta-2,4-diene drawn without stereochemistry. How many geometrical isomers does it have? Give an integer.

Both double bonds carry two different groups on each carbon (C2: $\mathrm{CH}{}_3$ and $\mathrm{H}$; C3 and C4: $\mathrm{H}$ and a carbon chain; C5: $\mathrm{C}{}_2\mathrm{H}{}_5$ and $\mathrm{H}$), so each can be $E$ or $Z$. The two ends of the molecule are different ($\mathrm{CH}{}_3$ versus $\mathrm{C}{}_2\mathrm{H}{}_5$), so no two of the combinations coincide. The number of geometrical isomers is $2^2=4$: $(2E,4E)$, $(2E,4Z)$, $(2Z,4E)$ and $(2Z,4Z)$.