Question

Find a cubic polynomial whose zero are 0,2/3,-2/3

Answer 1

let the zeros are
$\alpha \: \beta \: \gamma \\ \alpha + \beta + \gamma = 0 + \frac{2}{3} - \frac{2}{3} = 0 = \frac{ - b}{a} \\ \alpha \beta \gamma = 0( \frac{2}{3} )( - \frac{2}{3} ) = 0 = \frac{ - d}{a} \\ \alpha \beta + \beta \gamma + \alpha \gamma = 0 \times \frac{2}{3} + \frac{2}{3} \times \frac{ - 2}{3} + 0 \times \frac{ - 2}{3} \\ = - \frac{4}{9} = \frac{c}{a} \\ so \: the \: polynomial \: is\\ {x}^{3} - (\frac{ - b}{a}) {x}^{2} + \frac{c}{a} x - ( \frac{ - d}{a} ) = 0 \\ {x}^{3} - 0 {x}^{2} + ( \frac{ - 4}{9} )x - (0) = 0 \\ {x}^{3} - \frac{4}{9} x = 0 \\ 9 {x}^{3} - 4x = 0 \: \: \: is \: the \: answer$