Find a polynomial with integer coefficients that satisfies the given conditions.
$P$ has degree 2 and zeros $1+i \sqrt{2}$ and $1-i \sqrt{2}$

Step 1 ：The zeros of a polynomial are the values of x for which the polynomial equals zero. If a polynomial has integer coefficients and a complex zero, then the conjugate of that zero is also a zero of the polynomial. In this case, the zeros are \(1+i \sqrt{2}\) and \(1-i \sqrt{2}\), which are conjugates of each other.

Step 2 ：The polynomial with these zeros can be found by using the fact that a polynomial is the product of its factors. The factors corresponding to the zeros \(1+i \sqrt{2}\) and \(1-i \sqrt{2}\) are \((x - (1+i \sqrt{2}))\) and \((x - (1-i \sqrt{2}))\) respectively.

Step 3 ：So, the polynomial is \(P(x) = (x - (1+i \sqrt{2}))(x - (1-i \sqrt{2}))\).

Step 4 ：We can simplify this expression by multiplying out the brackets and combining like terms.

Step 5 ：The polynomial \(P(x) = x^2 - 2x + 3\) is the polynomial with integer coefficients that has the zeros \(1+i \sqrt{2}\) and \(1-i \sqrt{2}\).

Step 6 ：Final Answer: The polynomial is \(\boxed{x^2 - 2x + 3}\).