That cubic has no rational zeros.
. . I'll bet there's a typo . . .
How about: \(\displaystyle \;9x^3\,+\,39x^2\,+\,12x\:=\:0\) ?
. . Factor: \(\displaystyle \;3x(x^2\,+\,13x\,+\,4)\:=\:0\)
. . Factor: \(\displaystyle \;3x(x\,+\,4)(3x\,+\,1)\:=\:0\)
Therefore: \(\displaystyle \;x\,=\,0,\,-4,\,-\frac{1}{3}\)