## Intermediate Algebra (12th Edition)

$(x+2m+n)(x-2m-n)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $x^2-4m^2-4mn-n^2 ,$ group the last $3$ terms and factor the trinomial. Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the last $3$ terms, the given expression is equivalent to \begin{array}{l}\require{cancel} x^2-(4m^2+4mn+n^2) .\end{array} In the trinomial expression above, $a= 4 ,b= 4 ,\text{ and } c= 1 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 4(1)=4$ and whose sum is $b$ are $\left\{ 2,2 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} x^2-(4m^2+2mn+2mn+n^2) \\\\= x^2-[(4m^2+2mn)+(2mn+n^2)] \\\\= x^2-[2m(2m+n)+n(2m+n)] \\\\= x^2-[(2m+n)(2m+n)] \\\\= x^2-(2m+n)^2 .\end{array} The expressions $x^2$ and $(2mn+n)^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $x^2-(2m+n)^2 ,$ 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} (x)^2-(2m+n)^2 \\\\= [x+(2m+n)][x-(2m+n)] \\\\= (x+2m+n)(x-2m-n) .\end{array}