Answer
$p(x)=3x^4-9x^2+6$
Work Step by Step
Since $\sqrt{2}$ is a zero, to guarantee that the coefficients are integer it must be that $-\sqrt{2}$ is another zero.
Then,
$p(x)=A(x+1)(x-1)(x-\sqrt{2})(x+\sqrt{2})$
$p(x)=A(x^2-1)(x^2-2)$
$p(x)=A(x^4-3x^2+2)$
$p(x)=Ax^4-3Ax^2+2A$
In order that the constant term is $6$, it must be $2A=6\to A=3$.
Now, we get $p(x)=3x^4-9x^2+6$.
