Answer
$P(x)=3x^4-9x^2+6$
Work Step by Step
RECALL:
If c is a zero of a polynomial, then $x-c$ is a factor of the polynomial.
Since -1, 1, and $\sqrt2$ are zeros of the polynomial, and radical zeros always come in conjugate pairs, then
(x+1),(x-1), $(x-\sqrt2)$, and $(x+\sqrt2)$ are factors of the polynomial.
Thus, the polynomial of degree 4 with the given zeros is:
$\\P(x)=a(x+1)(x-1)(x-\sqrt2)(x+\sqrt2)
\\P(x)=a(x^2-1)(x^2-2)
\\P(x)=a(x^4-3x^2+2)$
where $a$ is a real number.
Since the constant term is 6, then $a$ must be:
$a(2)=6
\\a=\frac{6}{2}
\\a=3$
Thus, the polynomial with the given zeros whose constant term is 6 is:
$\\P(x)=3(x^4-3x^2+2)
\\P(x)=3x^4-9x^2+6$