Given that √2 is a zero …

Assume f(x) = {tex}6{x^3} + \sqrt 2 {x^2} - 10x - 4\sqrt 2{/tex}
If {tex}\sqrt 2{/tex} is the zero of f(x), then {tex}(x - \sqrt 2 ){/tex} will be a factor of f(x). So, by remainder theorem when f(x) is divided by {tex}(x - \sqrt 2 ){/tex}, the quotient comes out to be quadratic.
Now we divide  {tex}6{x^3} + \sqrt 2 {x^2} - 10x - 4\sqrt 2{/tex} by {tex}(x - \sqrt 2 ){/tex}.

{tex}\therefore \;f(x) = (x - \sqrt 2 )(6{x^2} + 7\sqrt 2 x + 4){/tex} (By Euclid’s division algorithm)
{tex}= (x - \sqrt 2 )(6{x^2} + 4\sqrt 2 x + 3\sqrt 2 x + 4){/tex} ( By factorization method )
For zeroes of f(x), put f(x) = 0
{tex}\therefore (x - \sqrt 2 )(6{x^2} + 4\sqrt 2 x + 3\sqrt 2 x + 4) = 0{/tex}
{tex}\Rightarrow \;(x - \sqrt 2 )[2x(3x + 2\sqrt 2 ){/tex} {tex}+ \sqrt 2 (3x + 2\sqrt 2 )] = 0{/tex}
{tex}\Rightarrow (x - \sqrt 2 )(3x + 2\sqrt 2 )(2x + \sqrt 2 ) = 0{/tex}
{tex}\Rightarrow x - \sqrt 2 = 0{/tex} or {tex}3x + 2\sqrt 2 = 0{/tex} or {tex}2x + \sqrt 2 = 0{/tex}
{tex}\Rightarrow x = \sqrt 2{/tex} or {tex}x = \frac{{ - 2\sqrt 2 }}{3}{/tex} or {tex}x = \frac { - \sqrt { 2 } } { 2 }{/tex}
So, other two roots are {tex}= \frac{{ - 2\sqrt 2 }}{3}{/tex} and {tex}\frac{{ - \sqrt 2 }}{2}{/tex}.