Question

Find a cubic polynomial having 1, 2, 3 as its zeroes.

Answer 1

Answer:

$x^{3}-6x^{2}+11x-6=0$

Step-by-step explanation:

let the roots of equation be

$\alpha = 1$       $\beta = 2$       $\gamma = 3$

$\alpha+\beta+\gamma = 6\\\alpha \beta + \beta \gamma +\alpha \gamma =2+6+3 = 11\\\alpha \beta \gamma =6$

hence the required equation is

$x^{3}-(\alpha +\beta +\gamma)x^{2}+(\alpha \beta +\beta \gamma + \alpha \gamma)x-\alpha \beta \gamma = 0$

$x^{3}-6x^{2}+11x-6=0$

Thus you got the answer !