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

1 Answer
Oct 3, 2015

Answer:

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

#ab^2-ab-72a = a(b-9)(b+8)#

Explanation:

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

#ab^2-ab-72a = a(b^2-b-72)#

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 xx 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 = (b-9)(b+8)#, giving us:

#ab^2-ab-72a = a(b-9)(b+8)#