College Algebra, Chapter 4, 4.5, Section 4.5, Problem 42

Find a polynomial $P(x)$ of degree 4 with interger coefficients and zeros $2i$ and $3i$.
Recall that if the polynomial function $P$ has real coefficient and if a $a + bi$ is a zero of $P$, then $a - bi$ is also a zero of $P$. In our case, we have zeros of $2i, -2i, 3i$ and $-3i$. Thus

$
\begin{equation}
\begin{aligned}
P(x) &= [x-2i][x-(-2i)][x-3i][x-(-3i)] && \text{Model}\\
\\
&= (x-2i)(x+2i)(x-3i)(x+3i) && \text{Simplify}\\
\\
&= (x^2 - 4i^2)(x^2 - 9i^2) && \text{Difference of squares}\\
\\
&= (x^2 +4)(x^2 + 9) && \text{Recall that } i^2 = -1\\
\\
&= x^4 + 9x^2 + 4x^2 + 36 && \text{Apply FOIL method}\\
\\
&= x^4 + 13x^2 + 36 && \text{Simplify}
\end{aligned}
\end{equation}
$