How do you factor #16j^3 + 44j^2 - 126j= 0#?
First, you factor out what's multiplying all your terms,
Now, we factor the parenthesis by finding its roots and turning them into factors by equaling them to zero.
I use the new AC method to factor trinomials (see Google ,Yahoo Search)
Compose factor pairs of a.c = -504-->(-12, 42)(-14, 36). This sum is 22 = b. Then p' = -14, and q' = 36.