## Algebra 1

$n^2-15n+56=(n-8)(n-7)$
$n^2-15n+56$ Factors of 56 | Sum of Factors $\ \ \ \ \ \ 1,56\ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ 57$ $\ \ \ \ \ \ 2,28\ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ 30$ $\ \ \ \ \ \ 4,14\ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ 18$ $\ \ \ \ \ \ 8,7\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ 15\checkmark$ Since the second term in the binomial is subtracted (or negative) and the third term is positive, the factors will both include subtraction. $n^2-15n+56=(n-8)(n-7)$ Use the FOIL method to check the answer. $(n-8)(n-7)=n^2-7x-8x+56=n^2-15n+56\checkmark$