Beginning Algebra With Applications, Chapter 8, 8.2, Section 8.2, Problem 58

Factor $c^2 - 3c - 180$

The factors must be of opposite signs


$
\begin{array}{c|c}
\text{Factors of -180} & \text{Sum} \\
\hline \\
-1,180 & 179 \\
1,-180 & -179 \\
-2,90 & 88 \\
2,-90 & -88 \\
-3,60 & 57 \\
3,-60 & -57 \\
-4,45 & 41 \\
4,-45 & -41 \\
-5,36 & 31 \\
5,-36 & -31 \\
-6,30 & 24 \\
6,-30 & -24 \\
-9,20 & 11 \\
9,-20 & -11 \\
-10,18 & 8 \\
10,-18 & -8 \\
-12,15 & 3 \\
12,-15 & -3
\end{array}

$


$c^2 - 3c - 180 = (c+12)(c-15) \qquad$ Write the factors of the trinomial