How do you factor completely #12x^2 - 10x - 8#?

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Answer 1 · 2016-04-07T20:03:35

#2(3x-4)(2x+1)#

Some times it is a matter of looking for common factors and experimenting.

Notice that the whole thing is exactly divisible by 2

#2(6x^2-5x-4)#

I have already experimented on paper!

Whole number factors of 6#-> {1,6}" , "{2,3}#

Whole number factors of 4#-> {1,4}" , "{2,2}#

Of these you are looking for a combination that will sum to #-5x#

#(3xx4)-(2xx1) = 10-># No good!

#(3xx1)-(2xx4)=-5-># Got it!

#2(3x-4)(2x+1)" "=" " 2(6x^2+3x-8x-4)#

#=2(6x^2-5x-4)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("The factors are: "2(3x-4)(2x+1))#