Question

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$2 {x}^{3} + y \\ \\ \\ \\ - \frac{1}{4} - xy$
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Answer 1

$(2 {x}^{3} + y)( - \frac{1}{4} - xy) = \frac{- 2 {x}^{3}}{4} - 2 {x}^{4} y - \frac{y}{4} - x {y}^{2} \\ \frac{ - 2 {x}^{3} }{4} - \frac{4}{4} \times 2 {x}^{4} y - \frac{y}{4} - \frac{4}{4} \times x {y}^{2} \\ \frac{- 2 {x}^{3} - 8 {x}^{4} y - y - 4x {y}^{2} }{4} = \frac{ - 2x( {x}^{2} + 4 {x}^{3} y + 2 {y}^{2} ) - y}{4} \\ \frac{ - 2x( {x }^{2} + y(4 {x}^{3} + 2y)) - y }{4}$