How do you factor the trinomial #6x^2-17x+12#?

1 Answer
Nov 2, 2016

#6x^2-17x+12 = (3x-4)(2x-3)#

Explanation:

Given:

#6x^2-17x+12#

Use an AC method:

Look for a pair of factors of #AC=6*12=72# with sum #B=17#

The pair #9, 8# works, in that #9*8=72# and #9+8=17#

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

#6x^2-17x+12 = 6x^2-9x-8x+12#

#color(white)(6x^2-17x+12) = (6x^2-9x)-(8x-12)#

#color(white)(6x^2-17x+12) = 3x(2x-3)-4(2x-3)#

#color(white)(6x^2-17x+12) = (3x-4)(2x-3)#