## College Algebra 7th Edition

(a) $\left\{-2i, 2i, -2, 2\right\}$. (b) $P(x) = (x-2)(x+2)(x-2i)(x+2i)$
$\bf{(a) \text{ Zeros}}$ Factor the polynomial completely to obtain: $P(x) = (x^2-4)(x^2+4) \\P(x) = ((x-2)(x+2)(x^2+4) \\P(x) = (x-2)(x+2)(x-2i)(x+2i)$ Equate each unique factor to zero then solve each equation to obtain: \begin{array}{cccccc} &x-2=0 &\text{or} &x+2=0 &\text{or} &x-2i=0 &\text{or} &x+2i=0 \\&x=2 &\text{or} &x=-2 &\text{or} &x=2i &\text{or} &x=-2i \end{array} Thus, the zeros of the function are: $\left\{-2i, 2i, -2, 2\right\}$. $\bf{(b) \text{ Completely Factored Form}}$ From part (a) above, the completely factored form of $P(x)$ is: $P(x) = (x-2)(x+2)(x-2i)(x+2i)$