[ x^2 - 5] / [ 3x^2 - 5x - 2 ] + [ x + 1] / [ 3x - 6]
Factor the denominators
[ x^2 - 5 ] / ( [ 3x + 1 ] [ x - 2] ) + [ x + 1 ] / [ 3 ( x - 2) ]
So....getting a common denominator, we have
[ 3 (x^2 - 5) + [ x + 1] [ 3x + 1] ] / [ 3 (x - 2) (3x + 1) ]
[ 3x^2 - 15 + 3x^2 + 3x + x + 1 ] / [ 3 (x - 2) (3x + 1 ) ]
[ 6x^2 + 4x - 14 ] / [ 3 (x - 2) (3x + 1 ) ]
[ 2 ( 3x^2 + 2x - 7 ) ] / [ 3 ( x - 2) (3x + 1) ] =
(6 x^2 + 4 x - 14) / (9 x^2 - 15 x - 6)
3x^2 - 5x - 2
Yeah....this can be difficult whenever the lead coefficient isn't a "1"
Let me show you a trick...it doesn't always work, but it sometines does
Write -5x as -6x + 1x and we have
3x^2 - 6x + 1x - 2 factor by grouping
3x ( x - 2) + 1 ( x - 2) the common factor is (x - 2)....so we have
( x - 2) (3x + 1) !!!
Here's another resource that might help..
http://www.coolmath.com/algebra/04-factoring/05-trinomials-undoing-FOIL-2-01