Question

Formula for finding cubic polynomial with their zeroes

Answer 1

Step-by-step explanation:

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Answer 2

General Cubic Equation

$f(x) = a {x}^{3} + b {x}^{2} + cx + d = 0$

$\implies g(z) = {z}^{2} + pz + q = 0$

$p = 2 {b}^{3} - 9abc + 27 {a}^{2} d \\ q = ({b}^{2} - 3ac) ^{3}$

Solutions:

$\therefore \underline{\: x _{1,2,3} = \frac{ - b \: + \: \omega_{k} \sqrt[3]{z_1} + { \omega_k}^{2} \sqrt[3]{z_2} }{3a}}$

where,

$\omega_k = 1, \frac{ - 1 + \sqrt{ - 3} }{2} , \frac{ - 1 - \sqrt{ - 3} }{2}$