To make this easier to factorise, multiply through by 3: 3x^3-2x^2+8x+3 (I assume me that 2/3x is supposed to be (2/3)x^2). Try x=1/3 as a zero: 3/27-2/9+8/3+3=(3-6+72+81)/27>0 so x=1/3 is not a solution; try x=-1/3: (-3-6-72+81)/27=0 so x=-1/3 is a zero. Use synthetic division:
-1/3 | 3 -2 8...3
.........3 -1 1..-3
.........3 -3 9 | 0
This gives us 3x^2-3x+9: factors so far: (x+1/3)(3x^2-3x+9)=(3x+1)(x^2-x+3). This does not factorise further except to produce complex roots. If we divide by 3 we get the original expression factorised: (x+1/3)(x^2-x+3).