College Algebra, Chapter 4, 4.5, Section 4.5, Problem 40

Find a polynomial $P(x)$ of degree 3 with interger coefficients and zeros $-3$ and $1 + i$.
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 $-3,1+i$ and $1-i$. Thus

$\begin{equation} \begin{aligned} P(x) &= [x-(-3)][x-(1+i)][x-(1-i)] && \text{Model}\\ \\ P(x) &= (x+3)[(x-1)-i][(x-1)+i] && \text{Regroup}\\ \\ P(x) &= (x+3) \left[ (x-1)^2 - i^2 \right] && \text{Difference of group}\\ \\ P(x) &= (x+3) \left[ x^2 - 2x +1 + 1\right] && \text{Expand, recall that } i^2 = -1\\ \\ P(x) &= (x+3) (x^2 - 2x + 2) && \text{Simplify}\\ \\ P(x) &= x^3 - 2x^2 + 2x + 3x^2 - 6x + 6 && \text{Expand}\\ \\ P(x) &= x^3 + x^2 - 4x + 6 && \text{Combine like terms} \end{aligned} \end{equation}$