College Algebra, Chapter 4, 4.5, Section 4.5, Problem 36

Find a polynomial $P(x)$ of degree that has integer coefficients and zeros $1 + \sqrt{2}i$ and $1 - \sqrt{2}i$.
To find the zeros of $P$, we set $x^6 + 16 x^3 + 64 = 0$, then,

$
\begin{equation}
\begin{aligned}
P(x) &= [x-(1+\sqrt{2}i)][x-(1-\sqrt{2}i)] && \text{Model}\\
\\
&= [(x-1)-\sqrt{2}i][(x-1)+\sqrt{2}i] && \text{Regroup}\\
\\
&= \left[ (x-1)^2 - 2i^2 \right] && \text{Difference of squares formula}\\
\\
&= \left[ x^2 - 2x + 1 - 2i^2 \right] && \text{Expand}\\
\\
&= x^2 -2x + 3 && \text{Simplify, recall that } i^2 = -1
\end{aligned}
\end{equation}
$