Answer
$(c+6)(c-3)$
Work Step by Step
$c^2+3c-18$ has no common monomial factor
$=(\ \ \ \ \ \ \ \ )(\ \ \ \ \ \ \ \ )$
$=(c\ \ \ \ \ \ )(c\ \ \ \ \ \ )$
$=(c+\ )(c-\ )$ The factors of -18 must have unlike signs.
$=(c+6)(c-3)$ 6 and -3 have a sum of 3 and a product of -18.
