How do you factor # 6y^2 + 27y - 15#?

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Answer 1 · 2016-09-19T06:17:02

#6y^2+27y-15 = 3(2y-1)(y+5)#

First note that all of the coefficients are divisible by #3#, so we can separate that out as a factor:

#6y^2+27y-15 = 3(2y^2+9y-5)#

We can factor the quadratic #2y^2+9y-5# using an AC method:

Find a pair of factors of #AC = 2*5 = 10# which differ by #B=9#

The pair #10, 1# works.

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

#2y^2+9y-5 = (2y^2+10y)-(y+5)#

#color(white)(2y^2+9y-5) = 2y(y+5)-1(y+5)#

#color(white)(2y^2+9y-5) = (2y-1)(y+5)#

Putting it together:

#6y^2+27y-15 = 3(2y-1)(y+5)#