## Intermediate Algebra (6th Edition)

$(x-4+y)(x-4-y)$
Since the first 3 terms of the given expression, $x^2-8x+16-y^2$, form a perfect square trinomial, then, \begin{array}{l} (x^2-8x+16)-y^2 \\\\= (x-4)^2-y^2 .\end{array} Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of two squares, then, \begin{array}{l} (x-4)^2-y^2 \\\\= [(x-4)+y][(x-4)-y] \\\\= (x-4+y)(x-4-y) .\end{array}