Algebra 2 help?

One way is to treat it as a quadratic in 'x'.
12x^2 + 32xy - 35y^2 = 0
-35 X 12 = -70 X 6 = 42 X -10.
Since 42 + (-10) = 32, required middle nomial is split
12x^2 + 42xy - 10xy + 35y^2 = 0
6x(2x + 7y) - 5y(2x + 7y) = 0
(6x -5y)(2x + 7y) = 0

The following are the pairs of factors of 12 1*12 2*6 3*4

The following are the pairs of factors of 35 1*35 5*7

With the equation

(a*x + b*y) * (c*x - d*y) (- because the -35 is negative)

we know that b * d = 35
a * c = 12
and a*d - b*y = 32

We look for the proper pairings

It takes time and arithmetic, but their aren't that many choices

The winner is:

(2x + 7y) * (6x - 5y) (my fourth guess)

First, look for a GCF. There is none in this problem. Then, factor. I use the guess & check method.
The answer is
(6x-5y)(2x+7y)

12XY((X/Y)+8/3-(35/12)(Y/X))

(6x -5y)(2x+7y)

12x^2 + 32xy - 35y^2
= 12x^2 + 42xy - 10xy - 35y^2
= (12x^2 + 42xy) - (10xy + 35y^2)
= 6x(2x + 7y) - 5y(2x + 7y)
= (2x + 7y)(6x - 5y)