Finding Zeros

x^3-3x^2+4x-12    factor and set to  0....we have.....

x^2 ( x - 3) + 4(x - 3)  = 0   take out the common factor ( x - 3)

(x - 3) (x^2  - 4)  = 0

(x - 3) ( x+ 2) ( x - 2)  =  0

And setting each factor to 0, the roots [ zeroes] are x = 3, -2 and 2

Hey Chris....   I think you made a mistake.   If you substitute those values you'll see

(x-3)(x^2   +   4) should be  third line

Hey Chris....   I think you made a mistake.   If you substitute those values you'll see

(x-3)(x^2   +   4) should be  third line  resulting in  3 and 2i as roots

..... and -2i