# Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 53

$8mn$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $(2m+n)^2-(2m-n)^2 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $(2m+n)^2$ and $(2m-n)^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $(2m+n)^2-(2m-n)^2 ,$ 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} [(2m+n)+(2m-n)][(2m+n)-(2m-n)] \\\\= (2m+n+2m-n)(2m+n-2m+n) \\\\= (4m)(2n) \\\\= 8mn .\end{array}

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