Given that x -√5 is a …

Let the given polynomials are f(x) = {tex}{x^3} - 3\sqrt 5 {x^2} + 13x - 3\sqrt 5{/tex}  and g(x) = {tex}(x - \sqrt 5 ){/tex}
∵ g(x) is a factor of f(x) so f(x) = q(x) {tex}(x - \sqrt 5 ){/tex}
Factor theorem, Euclid’s division algorithm.

But, f(x) = q(x) g(x)
{tex}\therefore f(x) = ({x^2} - 2\sqrt 5 x + 3)(x - \sqrt 5 ){/tex}
{tex}\Rightarrow f(x) = [{x^2} - \{ (\sqrt 5 + \sqrt 2 ) + (\sqrt 5 - \sqrt 2 )\} x{/tex}{tex} + (\sqrt 5 - \sqrt 2 )(\sqrt 5 + \sqrt 2 )][(x) - \sqrt 5 ]{/tex}
{tex}= \left[ {{x^2} - (\sqrt 5 + \sqrt 2 )x - (\sqrt 5 - \sqrt 2 )x} \right.{/tex}{tex}\left. { + (\sqrt 5 - \sqrt 2 )(\sqrt 5 + \sqrt 2 )][(x - \sqrt 5 )} \right]{/tex}
{tex}= x[ {x - (\sqrt 5 + \sqrt 2 )] - (\sqrt 5 - \sqrt 2 )}{/tex}{tex}{[x - (5 + \sqrt 2 )} ][x - \sqrt 5 ]{/tex}
For zeroes of f(x), f(x) = 0
{tex}\Rightarrow (x - \sqrt 5 - \sqrt 2 )(x - \sqrt 5 + \sqrt 2 )(x - \sqrt 5 ) = 0{/tex}
{tex}\Rightarrow (x - \sqrt 5 - \sqrt 2 ) = 0{/tex} or {tex}(x - \sqrt 5 + \sqrt 2 ) = 0{/tex} or {tex}(x - \sqrt 5 ) = 0{/tex}
{tex}\Rightarrow x = \sqrt 5 + \sqrt 2{/tex} or {tex}x = \sqrt 5 - \sqrt 2{/tex} or {tex}x = + \sqrt 5{/tex}
Hence the zeroes of given polynomoial are {tex}(\sqrt 5 + \sqrt 2 ),\;(\sqrt 5 - \sqrt 2 ){/tex} and {tex}\sqrt 5{/tex}.