A) expand the product $(x-1)(x-2)(x-3)$. b. the "cubic" $x^3+2x^2-11x-12$ has three roots. what is their sum?

A. (x-1)(x²-5x+6)=x³-5x²+6x-x²+5x-6=x³-6x²+11x-6.
b. Check if 1 is a zero: 1+2-11-12=-20. x-1 is not a factor. Try x=-1: -1+2+11-12=0, so x+1 is a factor.
Divide the cubic by the zero -1 using synthetic division:

-1 | 1 2 -11 -12
1 -1 -1 12
—————
1 1 -12 | 0 = x²+x-12 = (x-3)(x+4).

So the cubic is (x+1)(x-3)(x+4).