# How do you factor ab^2-ab-72a?

Oct 3, 2015

Separate out the common factor $a$, then find a pair of factors of $72$ that differ by $1$ to find:

$a {b}^{2} - a b - 72 a = a \left(b - 9\right) \left(b + 8\right)$

#### Explanation:

First notice that all the terms are multiple of $a$, so we can separate $a$ out as a factor to get:

$a {b}^{2} - a b - 72 a = a \left({b}^{2} - b - 72\right)$

Turning our attention to ${b}^{2} - b - 72$, we need to find a pair of numbers whose product is $72$ and whose difference is $1$ (the coefficient of the middle term).

There are several quick ways to find the pair $9 \times 8 = 72$.

One way is to notice that $72$ lies between the perfect squares $64 = {8}^{2}$ and $81 = {9}^{2}$.

Anyway, we can deduce that ${b}^{2} - b - 72 = \left(b - 9\right) \left(b + 8\right)$, giving us:

$a {b}^{2} - a b - 72 a = a \left(b - 9\right) \left(b + 8\right)$