## Intermediate Algebra: Connecting Concepts through Application

$6(a^2+4b^2)(a+2b)(a-2b)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $6a^4-96b^4 ,$ factor first the $GCF.$ Then use the factoring of the sum or difference of $2$ squares. $\bf{\text{Solution Details:}}$ The $GCF$ of the terms is $GCF= 6$ since it is the highest number that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 6(a^4-16b^4) .\end{array} The expressions $a^4$ and $16b^4$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $a^4-16b^4 ,$ 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} 6[(a^2)^2-(4b^2)^2] \\\\= 6(a^2+4b^2)(a^2-4b^2) .\end{array} The expressions $a^2$ and $4b^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $a^2-4b^2 ,$ 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} 6(a^2+4b^2)[(a)^2-(2b)^2] \\\\= 6(a^2+4b^2)(a+2b)(a-2b) .\end{array}