Answer
$a.\quad (n-3)(n+12)$
$b\quad (c+3)(c-7)$
Work Step by Step
a.
$-36 $ must have one positive and one negative factor
$\left[\begin{array}{lll}
\text{Factors of -36 } & \text{Sum of factors} & \\
1\text{ and }-36 & -35 & \\
-1\text{ and }36 & 35 & \\
2\text{ and }-18 & -16 & \\
-2\text{ and }18 & 16 & \\
3\text{ and }-12 & -9 & \\
-3\text{ and }12 & 9 & \text{...is what we need}
\end{array}\right]$
$n^{2}+9n-36=(n-3)(n+12)$
b.
$-21 $ must have one positive and one negative factor
$\left[\begin{array}{lll}
\text{Factors of -21 } & \text{Sum of factors} & \\
1\text{ and }-21 & -20 & \\
-1\text{ and }21 & 20 & \\
3\text{ and }-7 & -4 & \text{...is what we need} \\
-3\text{ and }7 & 4 &
\end{array}\right]$
$c^{2}-4c-21=(c+3)(c-7)$
