## College Algebra (11th Edition)

$(13y^2+1)(13y^2-1)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $169y^4-1 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Both $169y^4 \text{ and } 1$ 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} =(13y^2)^2-1^2 \\\\= (13y^2+1)(13y^2-1) .\end{array}