Using the Product Rule The given function is \(F(x)=8x^{4}(x^{2}-2x)\). Let \(u(x)=8x^{4}\) and \(v(x)=x^{2}-2x\). The derivative of \(u(x)\) is \(u^{\prime }(x)=\frac{d}{dx}(8x^{4})=8\cdot 4x^{4-1}=32x^{3}\). The derivative of \(v(x)\) is \(v^{\prime }(x)=\frac{d}{dx}(x^{2}-2x)=2x^{2-1}-2\cdot 1x^{1-1}=2x-2\). According to the Product Rule, \(F^{\prime }(x)=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)\). Substituting the expressions for \(u(x)\), \(v(x)\), \(u^{\prime }(x)\), and \(v^{\prime }(x)\) into the Product Rule formula, the derivative is \(F^{\prime }(x)=(32x^{3})(x^{2}-2x)+(8x^{4})(2x-2)\). Therefore, option B is the correct choice, and the blanks are filled as follows: The derivative is \(8x^{4}(2x-2)+(x^{2}-2x)(32x^{3})\). Part 2: Multiplying the Expressions The given function is \(F(x)=8x^{4}(x^{2}-2x)\). Multiplying the expressions, \(F(x)=8x^{4}\cdot x^{2}-8x^{4}\cdot 2x\). Simplifying the terms, \(F(x)=8x^{4+2}-16x^{4+1}=8x^{6}-16x^{5}\). Part 3: Taking the Derivative and Comparing Results The expanded form of the function is \(F(x)=8x^{6}-16x^{5}\). Taking the derivative of this expanded form, \(F^{\prime }(x)=\frac{d}{dx}(8x^{6}-16x^{5})\). Applying the power rule for differentiation, \(F^{\prime }(x)=8\cdot 6x^{6-1}-16\cdot 5x^{5-1}=48x^{5}-80x^{4}\). Now, simplifying the result from Part 1: \(F^{\prime }(x)=(32x^{3})(x^{2}-2x)+(8x^{4})(2x-2)\). Expanding the terms: \(F^{\prime }(x)=32x^{3}\cdot x^{2}-32x^{3}\cdot 2x+8x^{4}\cdot 2x-8x^{4}\cdot 2\). Simplifying the terms: \(F^{\prime }(x)=32x^{5}-64x^{4}+16x^{5}-16x^{4}\). Combining like terms: \(F^{\prime }(x)=(32x^{5}+16x^{5})+(-64x^{4}-16x^{4})=48x^{5}-80x^{4}\). Both approaches yield the same derivative, \(48x^{5}-80x^{4}\). Final Answer The final answer is \(\boxed{48x^{5}-80x^{4}}\)