Algebra 1: Common Core (15th Edition)

$a.\quad (n-3)(n+12)$ $b\quad (c+3)(c-7)$
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 $\left[\begin{array}{lll} \text{Factors of -21 } & \text{Sum of factors} & \\ 1\text{ and }-21 & -20 & \\ -1\text{ and }21 & 20 & \\ 3\text{ and }-7 & -4 & \text{...is what we need} \\ -3\text{ and }7 & 4 & \end{array}\right]$ $c^{2}-4c-21=(c+3)(c-7)$