College Algebra 7th Edition

(a) $\left\{-i, i \sqrt2, -\sqrt2 \right\}$. (b) $P(x) = (x-\sqrt2)(x+\sqrt2)(x-i)(x+i)$
$\bf{(a) \text{ Zeros}}$ Factor the polynomial completely to obtain: $P(x) = (x^2-2)(x^2+1) \\P(x) = (x-\sqrt2)(x+\sqrt2)(x-i)(x+i)$ Equate each unique factor to zero then solve each equation to obtain: \begin{array}{cccccc} &x-\sqrt2=0 &\text{or} &x+\sqrt2=0 &\text{or} &x-i=0 &\text{or} &x+i=0 \\&x=\sqrt2 &\text{or} &x=-\sqrt2 &\text{or} &x=i &\text{or} &x=-i \end{array} Thus, the zeros of the function are: $\left\{-i, i \sqrt2, -\sqrt2 \right\}$. $\bf{(b) \text{ Completely Factored Form}}$ From part (a) above, the completely factored form of $P(x)$ is: $P(x) = (x-\sqrt2)(x+\sqrt2)(x-i)(x+i)$