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

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Answer 1 · 2017-02-03T00:17:34

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

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)#