Question

Find a cubic polynomial whose zeros are −2, −3 and −1.

Answer 1

$\begin{array}{l}\text{Let}\alpha \text{=}-2\text{,}\beta \text{=}-3\text{and}\gamma \text{=}-1\\ \mathrm{Now},(\alpha +\beta +\gamma )=(-2-3-1)=-6\\ \mathrm{and}(\alpha \beta +\beta \gamma +\gamma \alpha )=(6+3+2)=11\\ \text{Also,}\alpha \beta \gamma \text{=}\left\{\right(-2)\times (-3)\times (-1\left)\right\}=(-6)\\ \therefore \text{Required polynomial =}{x}^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma \\ \text{i}\text{.e.,}{x}^3+6x^2+11x+6\\ \\ \end{array}$