Question

$If \alpha,\beta,\gamma be the zeros of the polynomial p(x) such that (\alpha+\beta+\gamma)=3,(\alpha\beta+\beta\gamma+\gamma\alpha)=-10 and a\beta\gamma=-24 then p(x)=?$

Answer 1

$</p><p></p><p>\bold{\underline{\sf Given:}} </p><p>\\ </p><p> \sf \rightarrow \quad \alpha + \beta + \gamma = 3 </p><p></p><p>\\ </p><p> \sf \rightarrow \quad \alpha \beta + \beta \gamma + \gamma \alpha = - 10 </p><p>\\ </p><p> \sf \rightarrow \quad \alpha \beta \gamma = -24</p><p></p><p>\\</p><p></p><p>\\ </p><p></p><p>\bold{\underline{\sf To Find:}} </p><p>\\ </p><p> \sf \rightarrow \quad p(x) = ?</p><p></p><p>\\ \\</p><p></p><p>\bold{\underline{\sf Solution:}} \\</p><p>\\ </p><p></p><p>\sf We \: Know, \\</p><p>\sf \; p(x) = x^{3} - ( \alpha + \beta + \gamma ) x^{2} + ( \alpha \beta + \beta \gamma + \gamma \alpha ) x - \alpha \beta \gamma \\ \\</p><p></p><p> \sf \rightarrow \quad x^{3} - ( 3 )x^{2} + (- 10)x - (-24)</p><p></p><p>\\ \\</p><p></p><p> \sf \rightarrow \quad x^{3} - 3x^{2} - 10x + 24</p><p></p><p>\\ \\</p><p></p><p>\sf \therefore Required \: polynomial \: p(x) \: is \: \\ \qquad \bold{ \sf x^{3} - 3x^{2} - 10x + 24}</p><p></p><p>$