How do you factor by grouping #12x^2-7x-5#?

Question

2161 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-06T13:00:15

The trick is to find how to split the middle #-7x# term.

The split comes from a pair of factors of #AC=12*5=60# whose difference is #B=7#. The pair #12, 5# works.

#12x^2-7x-5#

#= 12x^2-12x+5x-5#

#= (12x^2-12x)+(5x-5)#

#=12x(x-1)+5(x-1)#

#=(12x+5)(x-1)#

Answer 2 · 2015-06-06T22:02:01

When a + b + c = 0, you are advised to use the Shortcut.
One factor is (x - 1) and the other is (-c/a = 5/12).

y = (x - 1)( 12x + 5)