PreCalculus
posted by Ashley .
Hello, I have tried solving this using the rational roots theorem and none of the roots seem to be working. I'm trying to figure out where I went wrong.
3x^3 2x^2  7x  4 = 0

you are right, it does not factor over the rational numbers.
here is a webpage that solves cubics for you, just enter the constants
it gave me 1 irrational root and 2 complex roots. 
Thanks for the help. However, I don't see the webpage you were speaking of.

how silly of me
http://www.1728.com/cubic.htm 
Thanks!