College Algebra (10th Edition)

The zeros of the given function are: $\color{blue}{\left\{i, -i, -2i, 2i\right\}}$ The completely factored form of the given function is: $\color{blue}{P(x) = (x-i)(x+i)(x-2i)(x+2i)}$
Factor the trinomial to obtain: $P(x) = (x^2+4)(x^2+1)$ Equate each factor to zero then solve each equation to obtain: \begin{array}{ccc} &x^2+4=0 &\text{or} &x^2+1=0 \\&x^2=-4 &\text{or} &x^2=-1 \\&x=\pm\sqrt{-4} &\text{or} &x=\pm\sqrt{-1} \\&x=\pm\sqrt{4(-1)} &\text{or} &x=\pm i \\&x=\pm\sqrt{2^2(-1)} &\text{or} &x=\pm i \\&x=\pm2\sqrt{-1} &\text{or} &x=\pm i \\&x=\pm2i &\text{or} &x=\pm i \end{array} Thus, the zeros of the given function are: $\color{blue}{\left\{i, -i, -2i, 2i\right\}}$ The completely factored form of the given function is: $\color{blue}{P(x) = (x-i)(x+i)(x-2i)(x+2i)}$