Question

Find all zeroes of $2x^4-3x^3-3x^2+6x-2$ if $2$ zeroes $\sqrt{2}$ and $-\sqrt{2}$

Answer 1

$x=\pm \sqrt{2}$ are two roots

So, $(x+\sqrt{2})(x-\sqrt{2})=x^2-2$

And

$2x^4-3x^3-3x^2+6x-2=(x^2-2)(2x^2-3x+1)$

And for $2x^2-3x+1$,

$2x^2-3x+1=0$

$⟹(2x-1)(x-1)=0$

So, $x=1$ and$x=\frac{1}{2}$