I noted that n can be any real number(I suspect that n could even be
complex!!), for instance if m=2,t=2 and n=1/3 the equation T(A*n+B)=
(C*n+D)*(E*n+F) becomes

T(8*5+24) = (8*3+15)*({4/3}*25 + 20)
=T(64)= (39)*(160/3) = 65*32 = 64*65/2

Notice how the factors 13 and 5 of 65 conveniently become separated
between the two factors of T(64). This method could also work if we let
A*n +B be one less than any other number that needs factoring.

As an example of factoring 63 which is a Mersenne number, we can try
letting A*n+B = -64 by setting n = -11/15. Then my tri-equation becomes

T(-64)=-189/5 * -160/3 = 63*64/2 but this didn't factor the 63.

However, let T=2,M=3,n=1/2 then my tri-equation becomes
T(-64)= -36*(-56) which sucessfully factors the 63 by dividing by the factors of 64.

I believe this method has a lot of potential

Feel free to check my new group for more results. https://groups.yahoo.com/group/Figurate_Numbers