Question

Obtain all the zeroes of $x^4-3\sqrt{2}{x}^3+3x^2+3\sqrt{2}x-4$

Answer 1

One root $x=1$ and $x=-1$ can be found by hit and trial .

So f(x) =($x^2-1$)($x^2-3\sqrt{2}x+4$)=($x^2-1$)$(x-2\sqrt{2})(x-\sqrt{2})$

Thus other two roots are x=$2\sqrt{2},\sqrt{2}$