1. **Problem Statement:** We are given a polygon with four sides and three angles labeled as $2x$, $3x$, and an exterior angle $4x$. We need to find the value of $x$. 2. **Understanding the problem:** The sum of the interior angles of a quadrilateral is $360^\circ$. 3. **Key formula:** For any polygon, the sum of the interior angles is given by $$ (n-2) \times 180^\circ $$ where $n$ is the number of sides. For a quadrilateral, $n=4$, so the sum is $$ (4-2) \times 180^\circ = 360^\circ $$ 4. **Relationship between interior and exterior angles:** The exterior angle and its adjacent interior angle sum to $180^\circ$ because they form a linear pair. 5. **Set up equations:** Let the interior angles be $2x$ and $3x$, and the exterior angle be $4x$. The interior angle adjacent to the exterior angle $4x$ is $$180^\circ - 4x$$ 6. **Sum of interior angles:** The four interior angles are $2x$, $3x$, the angle adjacent to $4x$ which is $180^\circ - 4x$, and the fourth angle which we call $A$. 7. **Sum equation:** $$2x + 3x + (180 - 4x) + A = 360$$ 8. **Simplify:** $$2x + 3x - 4x + 180 + A = 360$$ $$ (2x + 3x - 4x) = x $$ So, $$ x + 180 + A = 360 $$ 9. **Solve for $A$:** $$ A = 360 - 180 - x = 180 - x $$ 10. **Sum of interior angles check:** Since $A$ is an interior angle, it must be positive and less than $180^\circ$. 11. **Use exterior angle property:** The exterior angle $4x$ equals the sum of the two opposite interior angles in a quadrilateral. The two opposite interior angles to $4x$ are $2x$ and $3x$. 12. **Set up exterior angle relation:** $$4x = 2x + 3x = 5x$$ 13. **Solve for $x$:** $$4x = 5x \implies 4x - 5x = 0 \implies -x = 0 \implies x = 0$$ 14. **Check for validity:** $x=0$ is not possible for angles. 15. **Re-examine the problem:** The exterior angle $4x$ is outside the polygon at the bottom-right corner. The interior angle adjacent to it is $180^\circ - 4x$. 16. **Sum of all interior angles:** $$2x + 3x + (180 - 4x) + A = 360$$ Simplify: $$x + 180 + A = 360 \implies A = 180 - x$$ 17. **Sum of interior angles is $360^\circ$, so all angles must be positive and less than $180^\circ$:** 18. **Since $A = 180 - x$, for $A$ to be positive:** $$180 - x > 0 \implies x < 180$$ 19. **Sum of angles around the point where $4x$ is an exterior angle:** The interior angle adjacent to $4x$ is $180 - 4x$. 20. **Sum of all interior angles:** $$2x + 3x + (180 - 4x) + (180 - x) = 360$$ Simplify: $$2x + 3x - 4x + 180 + 180 - x = 360$$ $$ (2x + 3x - 4x - x) + 360 = 360$$ $$0 + 360 = 360$$ 21. **This is true for all $x$, so we need another condition.** 22. **Use the fact that the exterior angle equals the sum of the two opposite interior angles:** $$4x = 2x + 3x = 5x$$ 23. **Solve:** $$4x = 5x \implies x = 0$$ which is invalid. 24. **Conclusion:** The problem as stated is inconsistent unless $x=0$, which is not possible for an angle. **Final answer:** No valid value of $x$ satisfies the given conditions unless $x=0$, which is not possible for an angle in a polygon.