1. **State the problem:** Simplify expressions for $A$, $B$, and $C$ given: $$A = \sqrt{150} - 2\sqrt{24} + \sqrt{36}$$ $$B = 3\sqrt{8} \times \frac{1}{2} \sqrt{2}$$ $$C = A(\sqrt{6} - B)$$ 2. **Simplify each term in $A$:** - $\sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6}$ - $2\sqrt{24} = 2 \times \sqrt{4 \times 6} = 2 \times 2\sqrt{6} = 4\sqrt{6}$ - $\sqrt{36} = 6$ So, $$A = 5\sqrt{6} - 4\sqrt{6} + 6 = (5-4)\sqrt{6} + 6 = \sqrt{6} + 6$$ 3. **Simplify $B$:** - $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$ - So, $B = 3 \times 2\sqrt{2} \times \frac{1}{2} \sqrt{2} = 3 \times \sqrt{2} \times \sqrt{2} = 3 \times 2 = 6$ 4. **Express $C = A(\sqrt{6} - B)$:** Substitute $A$ and $B$: $$C = (\sqrt{6} + 6)(\sqrt{6} - 6)$$ 5. **Expand $C$ using difference of squares:** $$C = (\sqrt{6})^2 - 6^2 = 6 - 36 = -30$$ **Final answers:** - $A = \sqrt{6} + 6$ - $B = 6$ - $C = -30$