## Intermediate Algebra for College Students (7th Edition)

$(a-10b)(a-8b)$
$(a+mb)(a+nb)=a^{2}+(m+n)ab+(mn)b^{2}$ $a^{2}-18ab+80b^{2}=$ Searching for two factors of $+80$ whose sum is $-18,$ we find$\qquad-10$ and $-8$ = $(a-10b)(a-8b)$