## Intermediate Algebra (12th Edition)

$3(4k^2+9)(2k+3)(2k-3)$
$\bf{\text{Solution Outline:}}$ 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}