How do you solve #6x^2 - 5x - 56 = 0# by factoring?

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Answer 1 · 2015-10-05T19:53:05

#(3x+8)(2x-7)=0#

#x=-8/3# or #x=7/2#

Problems like this take practice.
Undoing the FOIL method I try to look for factors where the product of the last terms is 56.

#(?+8)(?+7)=0#

The middle term is #-5x# so I take a guess that the first terms are
#3x# and #2x# but I have to account for the negative signs so I have

#(3x+8)(2x-7)=0#

#3x+8=0# or #2x-7=0#

#3x=-8#

#x=-8/3#

#2x=7#

#x=7/2#