Answer
$\left\{ -2, 2, -2i, 2i \right\}$
Work Step by Step
Using $a^2-b^2=(a+b)(a-b)$, then the factored form of the given expression, $
p^4-16=0
,$ is
\begin{array}{l}\require{cancel}
(p^2+4)(p^2-4)=0
.\end{array}
Equating each factor to zero, then,
\begin{array}{l}\require{cancel}
p^2+4=0
\\\\
p^2=-4
\\\\
p=\pm\sqrt{-4}
\\\\
p=\pm\sqrt{-1}\sqrt{4}
\\\\
p=\pm i\sqrt{(2)^2}
\\\\
p=\pm 2i
,\\\\\text{OR}\\\\
p^2-4=0
\\\\
p^2=4
\\\\
p=\pm\sqrt{4}
\\\\
p=\pm\sqrt{(2)^2}
\\\\
p=\pm2
.\end{array}
Hence, the solutions are $
\left\{ -2, 2, -2i, 2i \right\}
$.
