College Algebra, Chapter 4, 4.5, Section 4.5, Problem 44

Find a polynomial $P(x)$ that has degree 5 with integer coefficient and zeros $\displaystyle \frac{1}{2}, -1$ and $-i$, and leading coefficient 4; the zero $-1$ has multiplicity 2.
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 $\displaystyle \frac{1}{2}, -1, -i$ and $i$. Thus the required polynomial has the form.

$\begin{equation} \begin{aligned} P(x) &= a\left( x - \frac{1}{2} \right)[x - (-1)]^2 (x -i) [x - (-i)] && \text{Model}\\ \\ P(x) &= a\left( x - \frac{1}{2} \right) (x + 1)^2 (x-i)(x+i) && \text{Simplify}\\ \\ P(x) &= a\left( x - \frac{1}{2} \right) (x+1)^2 (x^2 -i^2) && \text{Difference of squares}\\ \\ P(x) &= a\left( x - \frac{1}{2} \right) (x+1)^2 (x^2 + 1) && \text{Recall that } i^2 = -1\\ \\\ P(x) &= a(x+1)^2 \left[ x^3 + x - \frac{1}{2}x^2 - \frac{1}{2} \right] && \text{Apply FOIL method}\\ \\ P(x) &= a \left[ x^2 + 2x + 1 \right] \left[ x^3 + x - \frac{1}{2}x^2 - \frac{1}{2} \right] && \text{Expand}\\ \\ P(x) &= a \left[ x^5 + x^3 - \frac{1}{2}x^4 - \frac{1}{2}x^2 + 2x^4 + 2x^2 - x^3 - x + x^3 + x - \frac{1}{2}x^2 - \frac{1}{2} \right] && \text{Expand}\\ \\ P(x) &= a \left[ x^5 + \frac{3}{2}x^4 + x^3 + x^2 - \frac{1}{2} \right] && \text{Simplify and combine like terms}\\ \\ P(x) &= 4 \left[ x^5 + \frac{3}{2}x^4 + x^3 + x^2 - \frac{1}{2} \right] && \text{Substitute } a = 4 \text{ to be the leading coefficient}\\ \\ P(x) &= 4x^5 + 6x^4 + 4x^3 + 4x^2 - 2 \end{aligned} \end{equation}$