Answer
$3(4k^2+9)(2k+3)(2k-3)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
48k^4-243
,$ factor first the $GCF.$ Then use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The $GCF$ of the terms in the given expression is $
3
,$ since it is the greatest expression that can divide all the terms evenly (no remainder.) Factoring the $GCF$ results to
\begin{array}{l}\require{cancel}
3(16k^4-81)
.\end{array}
The expressions $
16k^4
$ and $
81
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
16k^4-81
,$ 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}
3[(4k^2)^2-(9)^2]
\\\\=
3[(4k^2+9)(4k^2-9)]
\\\\=
3(4k^2+9)(4k^2-9)
.\end{array}
The expressions $
4k^2
$ and $
9
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
4k^2-9
,$ 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}
3(4k^2+9)[(2k)^2-(3)^2]
\\\\=
3(4k^2+9)[(2k+3)(2k-3)]
\\\\=
3(4k^2+9)(2k+3)(2k-3)
.\end{array}
