Question

Find the cubic evuotion whose roots are 1,2,-3​

Answer 1

       $\underline{\bold{Solution:}}$

       $\text{The given roots of the cubic equation are }1,2,-3$

$\implies \text{Sum of roots}(S_1) = 1 + 2 + (-3) = 1 + 2 -3 = 0$

$\implies \text{Sum of roots taken two at a time}(S_2) = (1)(2) + (2)(-3) + (-3)(1) = -7$

$\implies \text{Product of roots}(P) = (1)(2)(-3) = -6$

       $\text{We know that a cubic equation is of the form }x^{3} - (S_1)x^{2} + (S_2)x - (P)$

$\implies \text{The equation} = x^{3} - (0)x^{2} + (-7)x - (-6)$

$\implies \boxed{\text{The equation} = x^{3} -7x + 6}$