1. **State the problem:** Simplify the expression $$-2\sqrt{294m^{16}n^{7}}$$. 2. **Recall the rule for square roots:** $$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$$ and $$\sqrt{x^{2k}} = x^{k}$$ for even powers. 3. **Break down the expression inside the square root:** $$\sqrt{294m^{16}n^{7}} = \sqrt{294} \cdot \sqrt{m^{16}} \cdot \sqrt{n^{7}}$$ 4. **Simplify each part:** - $$\sqrt{294}$$: Factor 294 into prime factors: $$294 = 49 \times 6 = 7^{2} \times 6$$, so $$\sqrt{294} = \sqrt{7^{2} \times 6} = 7\sqrt{6}$$. - $$\sqrt{m^{16}} = m^{8}$$ because $$\sqrt{m^{16}} = m^{16/2} = m^{8}$$. - $$\sqrt{n^{7}} = \sqrt{n^{6} \times n} = \sqrt{n^{6}} \cdot \sqrt{n} = n^{3} \sqrt{n}$$. 5. **Combine all simplified parts:** $$7 \sqrt{6} \cdot m^{8} \cdot n^{3} \sqrt{n} = 7 m^{8} n^{3} \sqrt{6n}$$. 6. **Multiply by the outside coefficient -2:** $$-2 \times 7 m^{8} n^{3} \sqrt{6n} = -14 m^{8} n^{3} \sqrt{6n}$$. **Final answer:** $$\boxed{-14 m^{8} n^{3} \sqrt{6n}}$$