Answer
$f(x)=-2x^4+2x^3+4x+8$
Work Step by Step
If $\sqrt 2i$ is a zero of $f$ then the complex conjugate $-\sqrt 2i$ is also a zero.
$f(x)=a[(x-(-1)](x-2)(x-\sqrt 2i)[x-(-\sqrt 2i)]$
$f(x)=a(x+1)(x-2)(x-\sqrt 2i)(x+\sqrt 2i)$
$f(x)=a(x^2-2x+x-2)[x^2-(\sqrt 2i)^2]$
$f(x)=a(x^2-x-2)(x^2+2)$
$f(x)=a(x^4+2x^2-x^3-2x-2x^2-4)=a(x^4-x^3-2x-4)$
$f(1)=a(1^4-1^3-2(1)-4)=12$
$a(-6)=12$
$a=-2$
$f(x)=-2(x^4-x^3-2x-4)$
$f(x)=-2x^4+2x^3+4x+8$
