Find a cubic polynomial whose zeroes …

Let {tex}\alpha,\mathrm\beta\;\mathrm{and}\;\mathrm\gamma{/tex} be the zeroes of the given polynomial. 
Then, we have {tex}\alpha{/tex} = 3, {tex}\beta{/tex} = 5 and {tex}\gamma{/tex} = -2 
Hence
{tex}\alpha + \beta + \gamma{/tex} = 3 + 5 - 2 = 6    ...............(1)
{tex}\alpha \beta + \beta \gamma + \gamma \alpha{/tex} = 3(5) + 5(-2) + (-2)3 = 15 - 10 - 6 = -1    ................(2)
{tex}\alpha \beta \gamma{/tex} = 3(5)(-2) = -30     .............(3)
Now, a cubic polynomial whose zeros are  {tex}\alpha , \beta{/tex} and {tex}\mathrm\gamma{/tex} is equal to
p(x) = x³ - {tex}( \alpha + \beta + \gamma ) x ^ { 2 } + ( \alpha \beta + \beta y + \gamma \alpha ) x - \alpha \beta \gamma{/tex}
On substituting values from (1),(2) and (3) we get

{tex}\mathrm p(\mathrm x)=\mathrm x^3-(6)\mathrm x^2+(-1)\mathrm x-(-30){/tex}
= x³ - 6x² - x + 30