\( \begin{array}{ll}\text { Simplify Radicals } \\ \sqrt{125 v^{4}}= & \sqrt{6300}= \\ \sqrt{180}= \\ \sqrt[3]{8 y^{3}}= & \sqrt[4]{162 x^{5}}= \\ 1\end{array} \)

To simplify \( \sqrt{125 v^{4}} \), we can break it down: \( \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \) and \( \sqrt{v^{4}} = v^{2} \). Thus, \( \sqrt{125 v^{4}} = 5v^{2}\sqrt{5} \). For \( \sqrt{6300} \), we can factor it into \( 6300 = 100 \times 63 = 10^{2} \times 9 \times 7 \). Taking the square root gives us \( \sqrt{6300} = 10 \sqrt{63} = 10 \cdot 3\sqrt{7} = 30\sqrt{7} \). Next, simplifying \( \sqrt{180} \): we have \( \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \). For \( \sqrt[3]{8 y^{3}} \), we recognize that \( \sqrt[3]{8} = 2 \) and \( \sqrt[3]{y^{3}} = y \). Therefore, \( \sqrt[3]{8 y^{3}} = 2y \). Now for \( \sqrt[4]{162 x^{5}} \): we can break it down into \( 162 = 81 \cdot 2 = 3^{4} \cdot 2 \), which gives us \( \sqrt[4]{162} = 3\sqrt[4]{2} \). For \( x^{5} = x^{4} \cdot x = x \cdot \sqrt[4]{x} \), making our final result \( \sqrt[4]{162 x^{5}} = 3x\sqrt[4]{2x} \). So, summarizing the answers: \( \sqrt{125 v^{4}} = 5v^{2}\sqrt{5} \) \( \sqrt{6300} = 30\sqrt{7} \) \( \sqrt{180} = 6\sqrt{5} \) \( \sqrt[3]{8 y^{3}} = 2y \) \( \sqrt[4]{162 x^{5}} = 3x\sqrt[4]{2x} \) And the fun facts? You nailed them all!