1. Stating the problem: Simplify the expressions $\sqrt{294n^2}$ and $\sqrt{112x^3}$.\n\n2. Formula used: For any positive numbers $a$ and $b$, $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$. Also, $\sqrt{x^2} = |x|$.\n\n3. Simplify $\sqrt{294n^2}$:\n- Factor 294 into prime factors: $294 = 2 \times 3 \times 7^2$.\n- So, $\sqrt{294n^2} = \sqrt{2 \times 3 \times 7^2 \times n^2} = \sqrt{2} \times \sqrt{3} \times \sqrt{7^2} \times \sqrt{n^2}$.\n- Since $\sqrt{7^2} = 7$ and $\sqrt{n^2} = |n|$, we get $7|n|\sqrt{6}$.\n\n4. Simplify $\sqrt{112x^3}$:\n- Factor 112 into prime factors: $112 = 16 \times 7 = 2^4 \times 7$.\n- So, $\sqrt{112x^3} = \sqrt{2^4 \times 7 \times x^3} = \sqrt{2^4} \times \sqrt{7} \times \sqrt{x^3}$.\n- $\sqrt{2^4} = 2^2 = 4$.\n- $\sqrt{x^3} = \sqrt{x^2 \times x} = |x| \sqrt{x}$.\n- Therefore, $\sqrt{112x^3} = 4 |x| \sqrt{7x}$.\n\nFinal answers:\n$$\sqrt{294n^2} = 7 |n| \sqrt{6}$$\n$$\sqrt{112x^3} = 4 |x| \sqrt{7x}$$