Chapter 3, Polynomial and Rational Functions - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 330: 38

$P(x)=2x^{2}-4x+6$

Since $1-i\sqrt 2$ and $1+i\sqrt 2$ are zeros, this mean that P(x) have the following form: $P(x)=a[x-(1-i\sqrt 2)][x-(1+i\sqrt 2)]$ $P(x)=a[x-1+i\sqrt 2][x-1-i\sqrt 2]$ $P(x)=a[(x-1)^{2}-(i\sqrt 2)^{2}]$ $P(x)=a(x^{2}-2x+3)$ To make all all coefficients interger, we set $a=2$ and get $P(x)=2x^{2}-4x+6$