Find cubic polynomial whose zeroes are …

Let {tex}\alpha,\mathrm\beta\;\mathrm{and}\;\mathrm\gamma{/tex} be the zeroes of the given polynomial. 
Then, we have {tex}\alpha{/tex} = 2, {tex}\beta{/tex} = 1 and {tex}\gamma{/tex} = 1 
Hence
{tex}\alpha + \beta + \gamma{/tex} = 2 + 1 + 1 = 4    ...............(1)
{tex}\alpha \beta + \beta \gamma + \gamma \alpha{/tex} = 2(1) + 1(1) + 1(2) = 2 + 1 + 2 = 5    ................(2)
{tex}\alpha \beta \gamma{/tex} = 2(1)(1) = 2     .............(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-(4)\mathrm x^2+(5)\mathrm x-(2){/tex}
= x³ - 4x² + 5x - 2