## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$(x-6+a)(x-6-a)$
Grouping the first $3$ terms the given expression is equivalent to \begin{array}{l}\require{cancel} 36-12x+x^2-a^2 \\\\= (36-12x+x^2)-a^2 \\\\= (x^2-12x+36)-a^2 .\end{array} Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} (x^2-12x+36) \end{array} has $c= 36$ and $b= -12 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -6,-6 \right\}.$ Using these two numbers, the $\text{ expression }$ above is equivalent to \begin{array}{l}\require{cancel} (x-6)(x-6)-a^2 \\\\= (x-6)^2-a^2 .\end{array} The expressions $(x-6)^2$ and $a^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $(x-6)^2-a^2 ,$ 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-6)^2-(a)^2 \\\\= (x-6+a)(x-6-a) .\end{array}