College Algebra (11th Edition)

$(7m^4+3n)(7m^4-3n)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $49m^8-9n^2 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Both $49m^8 \text{ and } 9n^2$ are perfect squares (the square root is exact) and separated by a minus sign. 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} \\\\=(7m^4)^2-(3n)^2 \\\\=(7m^4+3n)(7m^4-3n) .\end{array}