Answer
$(k^2+4)(k+2)(k-2)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
k^4-16
,$ use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The expressions $
k^4
$ and $
16
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
k^4-16
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(k^2)^2-(4)^2
\\\\=
(k^2+4)(k^2-4)
.\end{array}
The expressions $
k^2
$ and $
4
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
k^2-4
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(k^2+4)[(k)^2-(2)^2]
\\\\=
(k^2+4)(k+2)(k-2)
.\end{array}