How do you factor completely #20y^2 + 3y - 2#?

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Answer 1 · 2017-01-09T00:47:34

#20y^2+3y-2 = (4y-1)(5y+2)#

Given:

#20y^2+3y-2#

Use an AC method:

Find a pair of factors of #AC=20*2 = 40# which differ by #B=3#

The pair #8, 5# works.

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

#20y^2+3y-2 = (20y^2+8y)-(5y+2)#

#color(white)(20y^2+3y-2) = 4y(5y+2)-1(5y+2)#

#color(white)(20y^2+3y-2) = (4y-1)(5y+2)#