Originally Posted by
HallsofIvy One of the things I am sure yuo have already learned is that if abc= 0 then at least one of a, b, or c must be 0.
Yes, you can factor $\displaystyle 2(x-3)(x+2)^2+ 2(x+2)(x-3)^3= 2(x+2)(x- 3)[(x+2)+ (x- 3)^2]= 2(x+2)(x-3)(x^2- 5x+ 11)= 0$.
So you know immediately that x+2= 0, or x- 3= 0, or $\displaystyle x^2- 5x+ 11= 0$ is 0. If x- 3= 0, then x= 3 so x= 3 is a root. If x+ 2= 0 then x= -2 (NOT 2) so x= -2 is a root. But there are two other roots- numbers that satisfy $\displaystyle x^2- 5x+ 11= 0$. That does not have any real number roots but you can find the complex roots by using the quadratic formula.