How do you factor #12y^(2)-4y-5#?

1 Answer
Feb 3, 2017

#12y^2-4y-5 = (2y+1)(6y-5)#

Explanation:

Use an AC method:

Given:

#12y^2-4y-5#

Look for a pair of factors of #AC = 12*5 = 60# which differ by #B=4#.

The pair #10, 6# works.

Use this pair to split the middle term and factor by grouping:

#12y^2-4y-5 = (12y^2-10y)+(6y-5)#

#color(white)(12y^2-4y-5) = 2y(6y-5)+1(6y-5)#

#color(white)(12y^2-4y-5) = (2y+1)(6y-5)#