1. **State the problem:** Simplify or analyze the expression $2x(x - i)^2(x - \sqrt{3i})^2$ where $i$ is the imaginary unit. 2. **Recall important rules:** - The imaginary unit $i$ satisfies $i^2 = -1$. - To simplify powers, expand binomials carefully. 3. **Expand $(x - i)^2$:** $$(x - i)^2 = x^2 - 2ix + i^2 = x^2 - 2ix - 1$$ 4. **Expand $(x - \sqrt{3i})^2$:** First, note $\sqrt{3i}$ is a complex number; let $a = \sqrt{3i}$. $$(x - a)^2 = x^2 - 2ax + a^2$$ Calculate $a^2 = 3i$. So, $$(x - \sqrt{3i})^2 = x^2 - 2x\sqrt{3i} + 3i$$ 5. **Rewrite the original expression:** $$2x (x^2 - 2ix - 1)(x^2 - 2x\sqrt{3i} + 3i)$$ 6. **Multiply the two quadratics:** Let $A = x^2 - 2ix - 1$, $B = x^2 - 2x\sqrt{3i} + 3i$. Calculate $AB$: $$AB = x^2(x^2 - 2x\sqrt{3i} + 3i) - 2ix(x^2 - 2x\sqrt{3i} + 3i) - 1(x^2 - 2x\sqrt{3i} + 3i)$$ Expand each term: - $x^2 \cdot x^2 = x^4$ - $x^2 \cdot (-2x\sqrt{3i}) = -2x^3\sqrt{3i}$ - $x^2 \cdot 3i = 3ix^2$ - $-2ix \cdot x^2 = -2ix^3$ - $-2ix \cdot (-2x\sqrt{3i}) = +4ix^2\sqrt{3i}$ - $-2ix \cdot 3i = -6i^2 x$ - $-1 \cdot x^2 = -x^2$ - $-1 \cdot (-2x\sqrt{3i}) = +2x\sqrt{3i}$ - $-1 \cdot 3i = -3i$ 7. **Simplify terms:** Recall $i^2 = -1$, so $-6i^2 x = -6(-1) x = 6x$. Group terms: - $x^4$ - $-2x^3\sqrt{3i} - 2ix^3$ - $3ix^2 + 4ix^2\sqrt{3i} - x^2$ - $6x + 2x\sqrt{3i}$ - $-3i$ 8. **Final expression:** $$2x \left[x^4 + (-2x^3\sqrt{3i} - 2ix^3) + (3ix^2 + 4ix^2\sqrt{3i} - x^2) + (6x + 2x\sqrt{3i}) - 3i \right]$$ This is the fully expanded form with complex terms. **Answer:** $$2x \left[x^4 - 2x^3\sqrt{3i} - 2ix^3 + 3ix^2 + 4ix^2\sqrt{3i} - x^2 + 6x + 2x\sqrt{3i} - 3i \right]$$