How do you factor #6x^2+x-1# ?

1 Answer
George C.
Jan 29, 2017

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

Explanation:

Here are a couple of methods (in no particular order):

Method 1

Note that:

#(ax+1)(bx-1) = abx^2+(b-a)x-1#

Comparing with:

#6x^2+x-1#

we want to find #a, b# such that #ab=6# and #b-a = 1#

The values #a=2#, #b=3# work, so we find:

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

#color(white)()#
Method 2 - Completing the square

To avoid much arithmetic with fractions, multiply first by #24 = 6*2^2# then divide by it at the end:

#24(6x^2+x-1) = 144x^2+24x-24#

#color(white)(24(6x^2+x-1)) = (12x)^2+2(12x)+1-25#

#color(white)(24(6x^2+x-1)) = (12x+1)^2-5^2#

#color(white)(24(6x^2+x-1)) = ((12x+1)-5)((12x+1)+5)#

#color(white)(24(6x^2+x-1)) = (12x-4)(12x+6)#

#color(white)(24(6x^2+x-1)) = (4(3x-1))(6(2x+1))#

#color(white)(24(6x^2+x-1)) = 24(3x-1)(2x+1)#

Divide both ends by #24# to get:

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

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