How do you factor #3x^2 + 5x - 12#?

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Answer 1 · 2015-05-30T19:00:24

Use a version of the AC Method.

#A=3#, #B=5#, #C=12#

Since the sign on the constant term is #-#, look for a pair of factors of #AC=36# whose difference is #B=5#

The pair #4#, #9# works.

Now use that pair to split the middle term, then factor by grouping:

#3x^2+5x-12#

#=3x^2-4x+9x-12#

#=(3x^2-4x)+(9x-12)#

#=x(3x-4)+3(3x-4)#

#=(x+3)(3x-4)#